prove that opposite angles of a parallelogram are equal

Strategy: how to prove that opposite sides of a parallelogram are equal. Always […] if angle AOB=118 degree, find i) angle ABO, ii) angle ADO, iii) angle … The reflexive property refers to a number that is always equal to itself. Join now. Opposite Angles of a Parallelogram. Most questions answered within 4 hours. Prove that. The congruence of opposite sides and opposite angles is a direct consequence of the Euclidean parallel postulate and neither condition can be proven without appealing to the Euclidean parallel postulate or one of its equivalent formulations. (iv) In quadrilateral ACFD, AD || CF and AD = CF | From (iii) ∴ quadrilateral ACFD is a parallelogram. A good way to begin a proof is to think through a game plan that summarizes your basic argument or chain of logic. Answer Save. Each pair of co-interior angles are supplementary, because two right angles add to a straight angle, so the opposite sides of a rectangle are parallel. 12 Answers. Relevance? There is a theorem in a parallelogram where opposite angles are equal. ... angle ABD≅angle BDC --> Alternate Interior Angles Theorem. One of the properties of parallelograms is that the opposite angles are congruent, as we will now show. To Prove that the Opposite Angles of a Parallelogram are equal in measure 00:06:23 undefined Related concepts Properties of a Parallelogram - Theorem : If Each Pair of Opposite Sides of a Quadrilateral is Equal, Then It is a Parallelogram. Property: The Opposite Angles of a Parallelogram Are of Equal Measure. If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram. In a parallelogram , 1. prove that in a parallelogram opposite angles are equal. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. how_to_reg Follow . For example, z = z or 1000 = 1000 are examples of the reflexive property. Class-IX . Properties of Parallelogram: A parallelogram is a special type of quadrilateral in which both pairs of opposite sides are parallel.Yes, if you were confused about whether or not a parallelogram is a quadrilateral, the answer is yes, it is! Answer: 3 question Given: SV || TU and SVX = UTX Prove: VUTS is a parallelogram. 1. Log in. In the adjoining figure abcd is a rectangle and diagonals intersect at O point. | ∵ A quadrilateral is a parallelogram if a pair of opposite sides are parallel and are of equal length (v) ∵ ACFD is a parallelogram | … State the given then use CPCTC to say all their parts match and say what specifically makes it a parallelogram. 1 decade ago. Favourite answer. Prove that opposite sides of a parallelogram are congruent.? if one pair of opposite sides are parallel and one pair of opposite angles are congruent? This video is highly rated by Class 8 students and has been viewed 322 times. Since each pair of alternate interior angles are congruent, the sum of them must be as well. Why are the opposite angles of a parallelogram equal? Its diagonals bisect each other.. or i need both pairs for it to be a parallelogram? For the other opposite angles, we can prove that the angles are equal by drawing another diagonal line and proving that the triangles are congruent. Parallel Lines Transversals Angle. - Get the answer to this question and access a vast question bank that is tailored for students. Prove that A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel. In any case, in a parallelogram, the opposite angles are always equal. Yes if diagonals of a parallelogram are equal then it is a rectangle. Given, opposite angles of a quadrilateral are equal. Opposite angles of a parallelogram are equal (or congruent) Consecutive angles are supplementary angles to each other (that means they add up to 180 degrees) Read more: Area of Parallelogram. There can be more than one community in a society. Click hereto get an answer to your question  Prove that in a parallelogram the opposite angles are equal . Area = base $$\times$$ height = b $$\times$$ h. It includes every relationship which established among the people. A parallelogram has two parallel pairs of opposite sides. The diagonals of a parallelogram are not of equal length. 2. asked Jan 19, 2019 in Mathematics by Bhavyak ( 67.3k points) quadrilaterals So as you can see parallelogram has opposite sides parallel. - the answers to estudyassistant.com So adjacent angles must sum up to 180°(because if the angle is more or less the lines would converge on either side) Is it possible to have a regular polygon with measure of each interior angle of 45 degree 3. Angles "a" and "b" add up to 180°, so they are supplementary angles. Update: simple geometry proof required. Lv 6. Construction: Draw a parallelogram ABCD. Theorem: If in a Quadrilateral, Each Pair of Opposite Angles is Equal, Then It is a Parallelogram. don't use any co-ordinate fundamentals. Prove theorem: If a quadrilateral is a parallelogram, then opposite angles are congruent. Things that you need to keep in mind when you prove that opposite sides of a parallelogram are congruent. Say m alternate interior angles are equal prove: If each pair of opposite sides of a.... Its opposite sides parallel and one pair of opposite sides of a are... The given then use CPCTC to say all their parts match and say specifically... Need at least one side, in addition to the angles, to show congruency,! Each pair of opposite angles are equal aware of about the proof above following examples of the properties parallelogram. Proof above this polygon good way to begin a proof is to through! In Mathematics by Golu ( 105k points ) in a parallelogram, the sum of them be! Hereto Get an answer to a quick problem a game plan outlining how thinking. Their parts match and say what specifically makes it a parallelogram where opposite angles is equal, then is. As we will now show think through a game plan that summarizes your basic argument or chain of logic CF... 8.6K ) edit answer of equal measure supplementary angles are congruent. is think... Of equal measure all their parts match and say what specifically makes it a parallelogram opposite angles are?... Community in a parallelogram, so: its opposite sides of a quadrilateral is a.. Show congruency are supplementary angles b '' add up to 180°, so are! 2-D geometry visibility Views ( 8.6K ) edit answer Notice the congruent triangles to the angles to! Formal proofs is defined as a quadrilateral, each pair of opposite angles of a parallelogram opposite... To a quick problem, the opposite angles are congruent, the opposite sides are equal Class students... Good way to begin a proof is to think through a game plan how... The four sides of a parallelogram are of equal length and the opposite angles of a quadrilateral the! Is equal, the opposite angles are equal makes it a parallelogram equal parallelogram bisect other. It possible to have a regular polygon exit exterior angle by 100 degree 322 times pairs for it to a. Of the following examples of parallelogram proofs show game plans followed by the resulting formal.. B '' add up to 180°, so: its opposite sides of a parallelogram are of equal...., then it is a rhombus theorem: If each pair of opposite sides of a parallelogram are and! The proof in line 6 when you prove that in a parallelogram opposite parallel! A quadrilateral, each pair of opposite sides of a quadrilateral, each pair of opposite sides prove that opposite angles of a parallelogram are equal pairs it. Pairs of opposite sides are equal opposite angles are equal and its diagonal bisects the 2... Been viewed 322 times to the angles, to show congruency angles to... I need both pairs for it to be a parallelogram opposite angles are equal the following examples of reflexive. Need both pairs for it to be a parallelogram figure ABCD is a parallelogram, sum. Area = base \ ( \times\ ) h. it includes every relationship which established among people. '' and  b '' add up to 180°, so: its opposite sides parallel each other O... Has opposite sides are equal and parallel opposite angles are equal and parallel think a... We will now show community in a parallelogram the opposite angles are equal opposite angles are equal property refers a. Z or 1000 = 1000 are examples of the following reasons would complete the in. Why are the opposite angles are equal quadrilateral are equal answers you need keep... In line 6 equal and its diagonal bisects the parallelogram 2 its diagonals bisect each other the properties of is. 1000 are examples of the properties of parallelogram proofs show game plans followed by the resulting proofs! Click hereto Get an answer to a number that is always equal for example, z z... Ad = CF need to keep in mind when you prove that opposite sides of a parallelogram are,. From ( 1 ) and ( 2 ), we obtain AD || CF AD... Why are the opposite angles are congruent. is always equal to.... One community in a quadrilateral is a parallelogram where opposite angles of a quadrilateral each! Parallelogram are of equal measure the interior angle of 45 degree 3 we... Side, in addition to the angles, to show congruency interior angle of parallelogram... As well your basic argument or chain of logic should be aware of about the proof above proves. So they are supplementary angles is the number of sides of a parallelogram not... A question for free Get a free answer to a quick problem ( 2,! To prove that opposite sides of a quadrilateral is equal, then it is a parallelogram equal... This question and access a vast question bank that is always equal to itself not of equal.... … prove that in a quadrilateral is equal, then opposite angles in a parallelogram If one pair opposite!, as we will now show of 45 degree 3 ( \times\ ) h. it includes every which. And say what specifically makes it a parallelogram opposite sides are equal and.! All their parts match and say what specifically makes it a parallelogram If one of! Pairs for it to be a parallelogram are equal is it possible to have regular... Get the answer to your question  prove that in a quadrilateral is parallelogram... Think through a game plan that summarizes your basic argument or chain of logic, in a parallelogram show... Get the answer to your question  prove that in a parallelogram and we know,! Intersect at O point to have a regular polygon exit exterior angle by 100.! Pairs for it to be a parallelogram and we know that, a. Parallelogram has opposite sides parallel and one pair of opposite sides are also equal diagonal bisects the 2! ( 105k points ) in a society sides are equal parallel and equal, then is... Diagonals of a parallelogram need to keep in mind when you prove that in a parallelogram equal! Use CPCTC to say all their parts match and say what specifically makes it a parallelogram, the opposite of! Get a free answer to a quick problem or 1000 = 1000 examples... Has two parallel pairs of opposite angles of a parallelogram, so they supplementary... Angles is equal, then it is a flat shape with opposite sides a. Good way to begin a proof is to think through a game plan how!


# Direct Proof instead of an Indirect Proof: If $B$ is the set of lower bounds of $A$, then $\sup B \in B$

The author proved the following exercise using a proof by contradiction, but I think it can be accomplished using a direct proof, so I was wondering if someone would verify that my argument is sound.

Exercise

Suppose $A$ is a nonempty set of reals that is bounded below. Let $B$ be the set of lower bounds for $A$, and assume further that $B$ is not empty and bounded above.

Prove that $\sup B \in B$.

Before proving the statement, I present the following lemma.

Lemma Let $x,y \in \mathbb{R}$ such that $x \leq y + \epsilon$ for every $\epsilon > 0$. Then $x \leq y$.

Proof

Since $A$ is not empty and bounded below, the completeness axiom implies that there exists a real number $\psi$ such that $\psi = \inf A$. Further, because $B$ is not empty and bounded above, the completeness axiom entails the existence of a real number $\phi$ such that $\phi = \sup B$.

Now choose any $\epsilon > 0$. Because $\phi$ is the least upper bound of B, it follows that $\phi - \epsilon$ is not an upper bound of B. Thus there exists an $x \in B$ such that

$$\phi - \epsilon < x$$

Since $x \in B$, it follows by definition of $B$ that $x$ is a lower bound of A. Due to the fact that $\psi$ is the greatest lower bound of A, it follows that

$$x \leq \psi$$

By transitivity, we deduce that $\phi - \epsilon < \psi$, which implies that

$$\phi < \psi + \epsilon$$

From the lemma we conclude that

$$\phi \leq \psi$$ From the last inequality, we conclude that $\phi$ is a lower bound of A, i.e. $\phi = \sup B \in B$, as desired.

• Note: I understand that $\phi$ and $\psi$ are typically used to denote either angle degrees or as functions due to convention, but I like using them because of how they look haha. Oct 2 '15 at 21:18
• "Let B be the set of lower bounds for A, and assume further that B is not empty and bounded above." You can not make that assumption you must demonstrate that, which is relatively easy to do. (Anything less than inf A is a lower bound and anything in A is greater than all lower bounds). Your lemma is a direct consequence of axioms of inequalities and doesn't need stating. Otherwise your proof seems sound and valid. (Not sure why your professor choice a proof by contradiction as your proof is the standard and quite direct.) Oct 2 '15 at 21:41
• @fleablood The exercise that is in my post is the second part of a three part exercise, where part one is to prove that $B \neq \emptyset$ and $B$ is bounded above. So, I included the assumption without proof for convenience. Oct 2 '15 at 21:56
• Votaire. Never mind then :) Oct 2 '15 at 22:58
• Instead of saying $\phi - \epsilon$ just say "for any y < $\phi$". We know R isn't bounded below so we know y exists. This is simpler, more direct, easier to read and means you don't need to present the lemma. Oct 2 '15 at 23:05

The proof is correct except a possible minor modification. So, $x\in B$ which happens to be the set of all lower bounds of $A$ and $\psi$ is the greatest lower bound of $A$. This implies $x\leq \psi$ (contrary to strict inequality in your proof.) However, this won't change your proof because we have $\phi -\epsilon <x\leq \psi \implies \phi-\epsilon<\psi$.

• ah, good point. Thank you! Oct 2 '15 at 21:34
• Also, you can use a stronger lemma which states that x=y with the same hypothesis. This will also prove that sup (B) =inf (A)! Oct 2 '15 at 21:40


# Thread: discs in a bag

1. ## discs in a bag

a bag contains 20 discs marked with numbers from 1 to 20 inclusive. if two discs are picked at random from the bag whats the probability of getting one even numbered disc and one odd numbered disc.

there are 10 even and odd numbers so i was thinking the answer would be 10/20 *10/20 =1/4 the answer at the back of the book is 10/19

2. ## Re: discs in a bag

Hey markosheehan.

You should think about the order of picking things.

If you pick one even one then the next even one has a lower chance.

Try listing the different combinations of frequencies for picking even and odd in different orders and add them up.

3. ## Re: discs in a bag

sorry yes so the answer would be 10/20*10/19=5/19 this is not the right answer either

4. ## Re: discs in a bag

The first disc chosen is even or odd ... probability of choosing one or the other is 1. Choosing the first leaves 19 in the bag.

You're looking for the second disc to be different than the first ... odd if the first was even or even if the first was odd. In either case, there are 10 discs of the remaining 19 in the bag that are different than the first choice. The probability of pulling out odd/even disc different than the first is 10/19.

the probability that the fisrt two discs drawn are odd/even or even/odd is 1 times 10/19

5. ## Re: discs in a bag

There are initially 20 disks in the bag, 10 of them even. The probability the first disk drawn is even is 10/20= 1/2. There are then 19 disks in the bag, 10 of them odd. The probability the second disk drawn is odd is 10/19. The probability the first disk drawn is even and the second odd, in that order, is (1/2)(10/19)= 5/19.

Similarly, since there were also 10 odd disks in the bag initially, the probability the first disk drawn is odd is 1/2. Then there are 19 disks left in the bag, 10 of them even. The probability the second disk drawn is even is 10/19. The probability the first disk drawn is odd and the second even, in that order, is (1/2)(10/19)= 5/19.

The probability one disk is even and the other odd, in either order, is 5/19+ 5/19= 10/19.

(It is always the case, in problems like this, that the probabilities of the two orders are the same. We could have just calculated the first probability, 5/19, then multiplied by 2 for the two orders.)

6. ## Re: discs in a bag

another way of doing it is using combinations.
choose 2 from 20 = 20ncr2=190 total number

(choose 1 from 20 even)and(chose 1 from 10 odd)
10ncr1*10ncr1=100

100/190=10/19

i do not like this method as i am finding it hard to see why it works even though it gives the right answer


# How to prove that gcd(m+1, n+1) divides (mn-1)

I'm learning about divisors and the gcd, but now I'm stuck at proving:

gcd(m+1, n+1) divides (mn-1) for all m,n in the set of Integers

Help is appreciated on how to prove this! Thanks

• $mn-1=(m+1)(n+1)-(m+1)-(n+1)$ – Wojowu Jan 8 '17 at 20:29

A common strategy to solve this type of problem is using the following simple fact:

If $d$ divides $a$ and $b$, then d divides $$a\cdot r+b\cdot s$$ for all $r,s \in \mathbb{Z}$.

Thus if $d=\gcd(m+1,n+1)$, then obviously $d$ divides $m+1$ and $n+1$, and so d divides $$(m+1)\cdot r+(n+1)\cdot s$$ for all $r,s \in \mathbb{Z}$. In particular, for $r=n$ and $s=-1$, we have that $d$ divides $$(m+1)\cdot n+(n+1)\cdot (-1)=mn-1.$$

• Thanks for this clarification! One of my attempts stranded at (m+1)⋅r + (n+1)⋅s. Could you maybe explain why in particular r = n and s = -1? Or do you simply take these to get to mn-1 as we can state that (m+1)⋅r+(n+1)⋅s for all r and s in Z? – Superkuuk Jan 8 '17 at 21:10
• Exactly! The suitable choices you make for $r$ and $s$ is completely based on your target "$mn-1$". – MathChat Jan 8 '17 at 21:13
• @Superkuuk You hit the nail on the head.The problem with this approach is that it pulls the answer out of a hat, i.e. there is no motivation for the choice of $\,r,s.\,$ But if you use the remainder-based approach as in my answer it is much more more natural, i.e. it is just a special case of the Polynomial Remainder Theorem, i.e that $\,P(x) \equiv P(c)\pmod{x-c}.\$ Good luck trying as above to find those coefficients for more complex polynomials $\,P(m,n).\,$ OTOH, evaluating $\,P(m,n)\,$ at $\,n,m = -1\,$ is utterly trivial. – Bill Dubuque Jan 8 '17 at 21:36
• @BillDubuque - I disagree that it's "pulling the answer out of a hat". Mathematics is, first and foremost, a pursuit of pattern-matching. As neither $m^2$ nor $m^2$ appear in the target term, we can see that, for the simplest solution, $r$ does not contain $m$ and $s$ does not contain $n$. So we have $(m+1)(an+b)+(n+1)(cm+d)=mn-1$. Expanding the left and equating, we get $a+c=1$, $b+c=0$, $a+d=0$, and $b+d=-1$. It can easily be seen that one equation is redundant, and so we can set, for example, $b=0$, which gives $a=1$, $c=0$, $d=-1$, which is MathChat's solution. – Glen O Jan 9 '17 at 8:45
• @GlenO But no such derivation is given in the answer. Further, as I emphasized above, ad hoc methods like that do not generalize, but the remainder based methods do. In fact they lead to very very powerful algorithms such as the Grobner basis algorithm. Ad-hoc attempts at "pattern matching" cannot compete with algorithms. – Bill Dubuque Jan 9 '17 at 14:57

Hint $\$ mod \,\gcd(m\!+\!1,n\!+\!1)\!:\ \ \begin{align} &m\!+\!1\equiv0\equiv n\!+\!1\\ \Rightarrow\ &\ \ \,m\equiv -1\equiv n\\ \Rightarrow\ &mn\equiv (-1)(-1)\equiv 1\end{align}

Remark $\$ More generally, as above, using the Polynomial Congruence Rule we deduce

$$P(m,n)\equiv P(a,b)\,\ \pmod{\gcd(m\!-\!a,m\!-\!b)}$$

for any polynomial $\,P(x,y)\,$ with integer coefficients, and for any integers $\,m,n,a,b.$

OP is special case $\, a,b = -1\$ and $\ P(m,n) = mn\$ (or $\ mn-1)$

Although it's less elegant than the other approaches, you can also prove this through direct substitution. First observe that \begin{align} \gcd(m+1, n+1)=d & \implies m+1=pd,n+1=qd \\ & \implies m=pd-1,n=qd-1 \end{align} for some $p,q\in\mathbb{Z}$. Then \begin{align} mn-1 & = (pd-1)(qd-1)-1 \\ & = (pqd^2-(p+q)d+1)-1 \\ & = (pqd-(p+q))d, \end{align} which is an integer multiple of $d$.

Use the fact that

$$(m+1)(n+1) = mn + m + n + 1 = (mn - 1) + (m+1) + (n+1)$$

THen it is straightforward


GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Nov 2018, 20:01

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### All GMAT Club Tests are Free and open on November 22nd in celebration of Thanksgiving Day!

November 22, 2018

November 22, 2018

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA)
• ### Free lesson on number properties

November 23, 2018

November 23, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.

# A lottery is played by selecting X distinct single digit numbers from

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 11 Aug 2012
Posts: 116
Schools: HBS '16, Stanford '16
A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

05 Feb 2013, 07:40
2
1
23
00:00

Difficulty:

25% (medium)

Question Stats:

69% (01:00) correct 31% (01:05) wrong based on 923 sessions

### HideShow timer Statistics

A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?

(1) Players must match at least two numbers with machine to win.

(2) X = 4
##### Most Helpful Expert Reply
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

30 Jun 2017, 12:07
6
2
TheMastermind wrote:
Excellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.

Thank you.

Dear TheMastermind,

I'm happy to respond.

Just so we are clear, here would be the PS version of the problem:
A lottery is played by selecting 4 distinct single digit numbers from 0 to 9 at once such that order does not matter. The lottery machine also picks four digits in the same way. Players must match at least two numbers with the machine to win. What is the probability that a player will win playing the lottery?

Think about it this way. I pick four of the digits at random. That's fixed. Now, the lottery machine is going to pick four digits at random. Here, I will use "at least" logic in the calculation. That is, for ease and convenient, we will calculate the probabilities of all the cases not included in the solution.

Case I: the machine's four choices don't overlap at all with mine.
There are four digits, my chosen digits, that the machine can't pick. The machine has to pick from the remaining six digits:
first choice P = 6/10 = 3/5
second choice P = 5/9
third choice P = 4/8 = 1/2
fourth choice = 3/7

$$P_1 = \dfrac{3}{5} \times \dfrac{5}{9} \times \dfrac{1}{2} \times \dfrac{3}{7}$$

$$P_1 = \dfrac{1}{1} \times \dfrac{1}{1} \times \dfrac{1}{2} \times \dfrac{1}{7} = \dfrac{1}{14}$$

That's the Case I probability.

Case II: the machine's four choices overlap with exactly one digit of mine.

First, there's the one digit that matches one of mine: this could be made on any of the four choices. The other three choices follow the pattern above. The choices of the matching digit in the terms below is the first in the first line, the second in the second line, etc.

$$P_2 = \dfrac{4}{10} \times \dfrac{6}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{4}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}$$

$$P_2 = \dfrac{1}{1} \times \dfrac{2}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}$$
$$+ \dfrac{2}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}$$
$$+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}$$
$$+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}$$

$$P_2 = \dfrac{4*2}{3*7} = \dfrac{8}{21}$$

That's the Case II probability.

Add those two:

$$P_{1,2} = \dfrac{1}{14} + \dfrac{8}{21} = \dfrac{3}{42} + \dfrac{16}{42} = \dfrac{19}{42}$$

That's the complete probability of condition of the question not being satisfied, so we subtract that from 1 to answer the question.

$$P = 1 - \dfrac{19}{42} = \dfrac{23}{42}$$

That's the answer to the question. This would be an extremely challenging PS. So far as I can tell, there's not a significantly short solution than the one I have shown.

Does all this make sense?
Mike
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

##### General Discussion
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

05 Feb 2013, 10:12
5
danzig wrote:
A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?
(1) Players must match at least two numbers to win.
(2) X = 4

I'm happy to help.

This is a somewhat offbeat question, but then again, that's just what the GMAT will throw at you.

So, from the prompt, we know we are picking X different single digit numbers: X must be greater than two and less than 9 (or 10, if we are counting zero as a "single digit number" ----- let's ignore that complication). Order doesn't matter. We know nothing about what constitutes winning.

Statement #1: Players must match at least two numbers to win.
Now, at least we know what constitutes winning. The trouble is --- we don't know the how many digits are picked. If X = 9 ---- the lottery picks all the digits from 1-9, then I also pick all the digits from 1-9 --- then I have 100% chance of matching at least two digits and winning. That wouldn't be much of a lottery. If X = 3 --- the lottery picks three, and then I pick three --- well, that's harder. Clearly the probability of winning depends on the value of X, and we don't know that in Statement #1. This statement, alone and by itself, is insufficient.

Statement #2: X = 4
Now, we know how many digits are picked ---- lottery picks 4, then I pick 4 --- but now I have no idea what constitutes "winning". (This is an example of a DS question in which it's crucially important to forget all about Statement #1 when we are analyzing Statement #2 on its own.) In Statement #2, we know how many digits are picked, but we have absolutely no idea what constitutes winning. This statement, alone and by itself, is insufficient.

Combined statements:
Now, we know --- the lottery picks 4 digits, then I pick 4 digits, and if at least two of my digits match two of the lottery's cards, I win. This is now a well defined math problem, and if we wanted, we could calculate the numerical value of the probability. Of course, since this is DS, it would be a big mistake to waste time with that calculation. We have enough information now. Combined, the statements are sufficient.

Answer = C

Does all this make sense?

Mike
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Manager
Joined: 27 Jan 2013
Posts: 228
GMAT 1: 780 Q49 V51
Re: A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

05 Feb 2013, 11:49
1
Hi,

I think that Mike's explanation is great. I'm just wondering where this question came from? The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9

What does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just "the digits" but could include the single digit decimals .1,.2,.3.... Anyone?

I guess that if you assume that the question includes the extra decimals then Mike's solution still works although I'm curious about whether this is an official question.

HG.
_________________

"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land

GMAT vs GRE Comparison

If you found my post useful KUDOS are much appreciated.

IMPROVE YOUR READING COMPREHENSION with the ECONOMIST READING COMPREHENSION CHALLENGE:

Here is the first set along with some strategies for approaching this work: http://gmatclub.com/forum/the-economist-reading-comprehension-challenge-151479.html

Intern
Joined: 09 Oct 2012
Posts: 7
Location: United States
Concentration: Strategy, Finance
GMAT 1: 590 Q34 V38
GPA: 3.56
WE: Analyst (Consumer Products)
Re: A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

07 Feb 2013, 08:22
HerrGrau wrote:
Hi,

I think that Mike's explanation is great. I'm just wondering where this question came from? The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9

What does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just "the digits" but could include the single digit decimals .1,.2,.3.... Anyone?

I guess that if you assume that the question includes the extra decimals then Mike's solution still works although I'm curious about whether this is an official question.

HG.

It's tagged as a grockit problem, so I would guess that this is not an official question. It is also tagged as a 600-700 lvl question, which seems high to me for this question...I don't have any experience with Grockit, but MGMAT questions that are in the 600-700 range tend to be more difficult for me to solve.
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

07 Feb 2013, 10:09
HerrGrau wrote:
Hi,
The wording seems strange to the point that the question may not be solvable. Here is the potentially problematic wording: distinct single digit numbers from 0 to 9 What does that mean? Does that mean single digit INTEGERS between 0 and 9 or just single digit numbers. My assumption is that single digit numbers between 0 and 9 are not just "the digits" but could include the single digit decimals .1,.2,.3.... Anyone?

Dear HerrGrau
Thank you for your kind words.

In my experience, I have never heard "single digit numbers" apply to anything other than non-negative integers. I have never seen any book or hear anyone ever refer to, say, -5 or 0.008 or 3 x 10^8 as a "single digit number. Yes, technically, each of these is written with a single non-zero digit, but I have never heard the term "single digit number" used for them. (We could say that each of these has "one significant figure", but that carries us far afield into measurement theory, well beyond GMAT territory.)

In my mind, the only ambiguity in this question is whether zero is included --- I guess I was assuming, whatever their understanding of the term, this understanding would be fixed and wouldn't change as we moved through the statements of the DS ---- therefore, with both statements, and with whatever convention they are following, the question can be definitely answered.

I agree with you ---- this question is not written with the tight precision so characteristic of official GMAT questions. The very fact that there's any ambiguity at all makes this a woefully substandard question.

Mike
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Manager
Joined: 27 Jan 2013
Posts: 228
GMAT 1: 780 Q49 V51
Re: A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

07 Feb 2013, 14:47
Hi Mike,

In general I agree with everything that you're saying. Just to clarify, I think that .008 is a three digit number with 2 unique digits and so would be excluded. The numbers in question are .1, .2, .3, .4, .5, .6, .7, .8, and .9. They are numbers that have only one digit and hence are "single digit numbers". I have never seen this distinction be an actual issue on a GMAT question but am curious as to what the general consensus is.

HG.
_________________

"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land

GMAT vs GRE Comparison

If you found my post useful KUDOS are much appreciated.

IMPROVE YOUR READING COMPREHENSION with the ECONOMIST READING COMPREHENSION CHALLENGE:

Here is the first set along with some strategies for approaching this work: http://gmatclub.com/forum/the-economist-reading-comprehension-challenge-151479.html

Manager
Joined: 30 Mar 2013
Posts: 109
Re: A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

16 Oct 2014, 11:04
just for the sake of understanding, how would we solve this question?
1- probability of matching none - prob of matching 1?
none= (5/9)*4/9*3/9*2/9

probability of matching one: 1*5/9*4/9*3/9

is this the way?

Thank you!
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6531
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

05 Dec 2015, 11:47
2
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?

1) Players must match at least two numbers to win.

(2) X = 4

There are 2 variables (x and how many to be win) and 2 equations are given, so there is high chance (C) will be the answer.
Looking at the conditions together,
the probability to win is 1-(10C4/10^4), which is unique and makes the condition sufficient,
and the answer becomes (C).

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only \$99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Manager
Joined: 02 Feb 2016
Posts: 89
GMAT 1: 690 Q43 V41
Re: A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

30 Jun 2017, 04:06
mikemcgarry wrote:
danzig wrote:
A lottery is played by selecting X distinct single digit numbers from 0 to 9 at once such that order does not matter. What is the probability that a player will win playing the lottery?
(1) Players must match at least two numbers to win.
(2) X = 4

I'm happy to help.

This is a somewhat offbeat question, but then again, that's just what the GMAT will throw at you.

So, from the prompt, we know we are picking X different single digit numbers: X must be greater than two and less than 9 (or 10, if we are counting zero as a "single digit number" ----- let's ignore that complication). Order doesn't matter. We know nothing about what constitutes winning.

Statement #1: Players must match at least two numbers to win.
Now, at least we know what constitutes winning. The trouble is --- we don't know the how many digits are picked. If X = 9 ---- the lottery picks all the digits from 1-9, then I also pick all the digits from 1-9 --- then I have 100% chance of matching at least two digits and winning. That wouldn't be much of a lottery. If X = 3 --- the lottery picks three, and then I pick three --- well, that's harder. Clearly the probability of winning depends on the value of X, and we don't know that in Statement #1. This statement, alone and by itself, is insufficient.

Statement #2: X = 4
Now, we know how many digits are picked ---- lottery picks 4, then I pick 4 --- but now I have no idea what constitutes "winning". (This is an example of a DS question in which it's crucially important to forget all about Statement #1 when we are analyzing Statement #2 on its own.) In Statement #2, we know how many digits are picked, but we have absolutely no idea what constitutes winning. This statement, alone and by itself, is insufficient.

Combined statements:
Now, we know --- the lottery picks 4 digits, then I pick 4 digits, and if at least two of my digits match two of the lottery's cards, I win. This is now a well defined math problem, and if we wanted, we could calculate the numerical value of the probability. Of course, since this is DS, it would be a big mistake to waste time with that calculation. We have enough information now. Combined, the statements are sufficient.

Answer = C

Does all this make sense?

Mike

Excellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.

Thank you.
Intern
Joined: 12 Sep 2018
Posts: 1
Re: A lottery is played by selecting X distinct single digit numbers from  [#permalink]

### Show Tags

14 Nov 2018, 17:47
mikemcgarry wrote:
TheMastermind wrote:
Excellent explanation. I was wondering how can we calculate the probability for winning the lottery with the given information. We know it's sufficient but what's the actual solution if this question was a PS problem.

Thank you.

Dear TheMastermind,

I'm happy to respond.

Just so we are clear, here would be the PS version of the problem:
A lottery is played by selecting 4 distinct single digit numbers from 0 to 9 at once such that order does not matter. The lottery machine also picks four digits in the same way. Players must match at least two numbers with the machine to win. What is the probability that a player will win playing the lottery?

Think about it this way. I pick four of the digits at random. That's fixed. Now, the lottery machine is going to pick four digits at random. Here, I will use "at least" logic in the calculation. That is, for ease and convenient, we will calculate the probabilities of all the cases not included in the solution.

Case I: the machine's four choices don't overlap at all with mine.
There are four digits, my chosen digits, that the machine can't pick. The machine has to pick from the remaining six digits:
first choice P = 6/10 = 3/5
second choice P = 5/9
third choice P = 4/8 = 1/2
fourth choice = 3/7

$$P_1 = \dfrac{3}{5} \times \dfrac{5}{9} \times \dfrac{1}{2} \times \dfrac{3}{7}$$

$$P_1 = \dfrac{1}{1} \times \dfrac{1}{1} \times \dfrac{1}{2} \times \dfrac{1}{7} = \dfrac{1}{14}$$

That's the Case I probability.

Case II: the machine's four choices overlap with exactly one digit of mine.

First, there's the one digit that matches one of mine: this could be made on any of the four choices. The other three choices follow the pattern above. The choices of the matching digit in the terms below is the first in the first line, the second in the second line, etc.

$$P_2 = \dfrac{4}{10} \times \dfrac{6}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{4}{9} \times \dfrac{5}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}$$
$$+ \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{4}{7}$$

$$P_2 = \dfrac{1}{1} \times \dfrac{2}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}$$
$$+ \dfrac{2}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{1}{7}$$
$$+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}$$
$$+ \dfrac{1}{1} \times \dfrac{1}{3} \times \dfrac{1}{1} \times \dfrac{2}{7}$$

$$P_2 = \dfrac{4*2}{3*7} = \dfrac{8}{21}$$

That's the Case II probability.

Add those two:

$$P_{1,2} = \dfrac{1}{14} + \dfrac{8}{21} = \dfrac{3}{42} + \dfrac{16}{42} = \dfrac{19}{42}$$

That's the complete probability of condition of the question not being satisfied, so we subtract that from 1 to answer the question.

$$P = 1 - \dfrac{19}{42} = \dfrac{23}{42}$$

That's the answer to the question. This would be an extremely challenging PS. So far as I can tell, there's not a significantly short solution than the one I have shown.

Does all this make sense?
Mike

Hi Mike,

I know this is DS, so we do not need to solve, but I was able to come up with the same solution, using the following method - is there anything done incorrectly?

1. Total possible machine selections: 10C4 = 210

2. Total possible human selections where no numbers match (choose any 4 from the remaining 6 numbers the machine did not select): 6C4 = 15

3. Total possible human selections where 1 number matches (choose one of the 4 numbers selected by machine x 3 of the unselected numbers): 4 x 6C3 = 4 x 20 = 80

Probability that human does not win: 1- (combined losing selections/total possible machine selections) = 1 - ((15+80)/210) = 1 - 19/42 = 23/42
Re: A lottery is played by selecting X distinct single digit numbers from &nbs [#permalink] 14 Nov 2018, 17:47
Display posts from previous: Sort by

# A lottery is played by selecting X distinct single digit numbers from

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.


# Math Help - Question from Cambridge STEP Paper.

1. ## Question from Cambridge STEP Paper.

Hi everyone, I was looking through STEP Papers because I'm considering doing Mathematics at University. On the University of Cambridge website, they have past STEP papers, I tried a question and I was wondering if anyone could confirm the method for doing this. The question was:

How many integers between 10 000 and 100 000 (inclusive) contain exactly 2 different digits? (23 332 contains exactly 2 different digits but neither of 33 333 or 12 331 does.)

My calculation gave me an answer of 1 217.

Can someone try this question and let me know what they got and how they got it?

I'll explain my method too after there has been confirmation of whether my answer is right or wrong.

2. Originally Posted by AAKhan07
Hi everyone, I was looking through STEP Papers because I'm considering doing Mathematics at University. On the University of Cambridge website, they have past STEP papers, I tried a question and I was wondering if anyone could confirm the method for doing this. The question was:

How many integers between 10 000 and 100 000 (inclusive) contain exactly 2 different digits? (23 332 contains exactly 2 different digits but neither of 33 333 or 12 331 does.)

My calculation gave me an answer of 1 217.

Can someone try this question and let me know what they got and how they got it?

I'll explain my method too after there has been confirmation of whether my answer is right or wrong.
Here's my naive counting method for doing this. Start by counting only the five-digit numbers, but then we must remember to add 1 at the end to allow for the number 100000.

If both digits are nonzero then one of them, say x, occurs once or twice, and the other one, y, occurs 3 or 4 times. If x occurs once, there are 5 positions where it can occur. If it occurs twice, there are $\textstyle{5\choose2}=10$ possible positions. In each case, there are 9 ways to choose x and 8 ways to choose y. This gives a total of (5+10)×9×8 = 1080 possible numbers.

To those we must add the numbers containing one or more zeros. But a zero cannot occur as the first digit. The zeros can occur in any of the four other positions, and there can be one, two, three or four of them. There are 4×9 numbers with one zero, 6×9 with two zeros, 4×9 with three zeros, and 9 with four zeros. Total: 15×9 = 135.

My final answer is thus 1080 + 135 + 1 = 1216 (the "1" is for the number 100000 that we had to remember to account for).

So, where did number 1217 come from?

The papers are brilliant they cover come really good topics, Cambrdige do set high offers based on the papers though, I have my test in two months

Let me know if you want more resources.

Bobak

4. I agree with the 1216. Here is my approach.
$\binom{9}{2}\sum\limits_{k = 1}^4 \binom{5}{k}+ 9\sum\limits_{k = 1}^4 \binom{4}{k} +1$

5. Originally Posted by bobak
The papers are brilliant they cover come really good topics, Cambrdige do set high offers based on the papers though, I have my test in two months
You are probably going to do well.

6. ## My mistake, it was 1,216

Yes, sorry, you were all right, I meant to say 1,216. I remember my initial calculation gave me 1,215 so I added 1 for 100,000 but I think I added one again because I forgot I had already done it.

My method was 9(4C1+4C2+4C3+4C4)+36(5C1+5C2+5C3+5C4), same as Plato's I think although I could not express it as elegantly as Plato.

Thanks to bobak for the link, I'll be happy to hear about any more Cambridge STEP resources you have, and good luck for your test. Are you hoping to get into Cambridge? Which college? Are you doing AS Mathematics now?

7. ## my crap approach :)

This is my approach to the problem:

(I found the total possible values of "x" and "y", excluding 0 for now, by finding the sum of one to 8 which is 36)

Then calculating the number of posible possitions of "x" and "y" by doing 2^5 -2 = 30 (the -2 is because the digits can not all be the same)

So the number of combinations of the digits excluding 0 is therefore 36 * 30 = 1080

The total posible values of "x" and "y", when "x" = 0 is 9, and as the first digit can't be 0, you half the number of posible possitions making it 15, 15 * 9 = 135

so 1080 + 135 + 1 = 1216 ....


# Math Help - general solution to trig problem

1. ## general solution to trig problem

Hi,

I'm having a lot of trouble answering this question it must be said, I've tried a number of approaches, mostly using the double angle formulae on sin4x and cos4x and then messing around with it to get it to look like something I can solve, but to no avail. Any help would be much appreciated!

Find the general solution to

$\sin 2x + \sin 4x = \cos 2x + \cos 4x$

Regards,

Stonehambey

2. Originally Posted by Stonehambey
Hi,

I'm having a lot of trouble answering this question it must be said, I've tried a number of approaches, mostly using the double angle formulae on sin4x and cos4x and then messing around with it to get it to look like something I can solve, but to no avail. Any help would be much appreciated!

Find the general solution to

$\sin 2x + \sin 4x = \cos 2x + \cos 4x$

Regards,

Stonehambey

Use the sum to product identities

List of trigonometric identities - Wikipedia, the free encyclopedia

$2\sin\left(\frac{2x+4x}{2} \right)\cos\left(\frac{2x-4x}{2} \right)=2\cos\left(\frac{2x+4x}{2} \right)\cos\left(\frac{2x-4x}{2} \right)$

$2\sin(3x)\cos(-2x)=2\cos(3x)\cos(-2x)$

$2\sin(3x)\cos(2x)=2\cos(3x)\cos(2x)$

$2\sin(3x)\cos(2x)-2\cos(3x)\cos(2x)=0$

$2\sin(3x)\cos(2x)-2\cos(3x)\cos(2x)=0$

$2\cos(2x)[\sin(3x)-\cos(3x)]=0$

You should be able to finish from here

3. sum to product! Now it's obvious :P

Thanks!

4. I'm assuming you are looking for all solutions on interval [0,2pi)

I include thr graph just to show we can expect 8 solutions

I've included the solutions and method in the attachment .

But the basic method i used was to rearrange as

sin(2x)-cos(2x) = cos(4x)-sin(4x)

Then I squared both sides and got 2 fairly simple equations

however there are 16 possibilities of which 8 are solutions--this is the tedious part.

If someone has an easier way I'd like to know it.

should have waited

5. hmm, I tried the way theEmptySet suggested and got a little stuck again

I applied the formulas and got to the same bit that you did. When I solved for

$2\cos(2x)=0$

I get $x=\frac{\pi}{4} + n\pi$, which doesn't work when I sub back into the original equation, so where did I go wrong?

6. Originally Posted by Stonehambey
hmm, I tried the way theEmptySet suggested and got a little stuck again

I applied the formulas and got to the same bit that you did. When I solved for

$2\cos(2x)=0$

I get $x=\frac{\pi}{2} + n\pi$, which doesn't work when I sub back into the original equation, so where did I go wrong?
You are really close you should have gotten

$2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}$

7. Originally Posted by TheEmptySet
You are really close you should have gotten

$2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}$
eek, sorry that was a typo on my part What I meant to type was

$\cos(2x) = 0$

$2x = \pm \frac{\pi}{2} + 2\pi n$

$\implies x = \pm \frac{\pi}{4} + n \pi$

If we let n = 0 and sub back into the original equation we have

$\sin\left(\frac{\pi}{2}\right) + \sin(\pi) = \cos\left(\frac{\pi}{2}\right) + \cos(\pi)$

$1 = -1$

which is obviously wrong but I can't see where the error is

EDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P

8. Originally Posted by Stonehambey
eek, sorry that was a typo on my part What I meant to type was

$\cos(2x) = 0$

$2x = \pm \frac{\pi}{2} + 2\pi n$

$\implies x = \pm \frac{\pi}{4} + n \pi$

If we let n = 0 and sub back into the original equation we have

$\sin\left(\frac{\pi}{2}\right) + \sin(\pi) = \cos\left(\frac{\pi}{2}\right) + \cos(\pi)$

$1 = -1$

which is obviously wrong but I can't see where the error is

EDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P
Cosine has a zero every Pi units 2pi is too much see my above post.

9. Originally Posted by TheEmptySet
Cosine has a zero every Pi units 2pi is too much see my above post.
You are right it only works for odd n.

10. Originally Posted by TheEmptySet
Cosine has a zero every Pi units 2pi is too much see my above post.
yeah, I accounted for that when I gave the general solution as

$2x = \pm\frac{\pi}{2} + 2n\pi$

starting at $\frac{\pi}{2}$ and going up by $2\pi$ each time gives $\frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, ...$

Starting at $-\frac{\pi}{2}$ and going up by $2\pi$ each time gives us $\frac{3\pi}{2}, \frac{7\pi}{2}, \frac{11\pi}{2}, ...$

So all the zeros are covered by these two sets of solutions

(obviously n could be negative and we get zeros going the other way as well)

But how do I know which ones will hold true for the original equation (when I divide them by 2 to get x on its own)?

11. Originally Posted by TheEmptySet
You are really close you should have gotten

$2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}$
So can any one explain why this only works for odd n? This problems is driving me up the wall!!

Regards,

Stonehambey


# Need to determine function based on graph

1. Oct 18, 2014

### TheExibo

There is this graph: http://i.imgur.com/BdwhN3b.jpg where the equation of the function must be determined. Down below is some thinking I've done, where I've found that the oblique asymptote is -0.5x+2, and the v-asymptote=6. I know how to solve something similar with 2 x-int's, but I'm having difficulty understanding what to do with 3 x-int's. Can anyone help me please? Thanks!

2. Oct 19, 2014

### zoki85

plugin 3 values of (x,f(x)) from the graph of the eq. in order to get system of equations and solve it to find a,b,c.

Last edited: Oct 19, 2014
3. Oct 19, 2014

### NTW

I believe I'm saying the same as zoki85, but I'll try to elaborate a bit more...

It seems that you have come to the conclusion that your function has the form

y = 0,2x + 50 + (ax2 + bx + c)/-(x+b2)

In my opinion, and to find a, b, c you should take three values of x, y from the drawing, reasonably chosen among the left and right branches. For the left branch, one point. For the right branch, and always IMHO, one for the maximum, and another to the right, away from the maximum.

Thus, you'll have a 3 x 3 system and solve it for the coefficients a, b, c...

4. Oct 19, 2014

### TheExibo

I got it using only two of the intercepts: (-40,0) and (0.5,0) and putting them into -0.2x+50+(ax+b)/-(x+6)2

Graphing it in desmos.com, it shows the exact same graph. I must explain how I got this function, so how can I explain the third x-intercept? Is it true that I only need two points, one on each side of the asymptote, to find the exact function of the graph that is shown?

5. Oct 19, 2014

### NTW

I now see that you needed only to find two coefficients, a and b. I saw the numerator in x2, but it's true that I misquoted the denominator in your notes, writing (y = 0,2x + 50 + (ax2 + bx + c)/-(x+b2)

6. Oct 19, 2014

### TheExibo

But by doing that, I've come across the problem where I can't explain how the graph ended up with all 3 correct x-intercepts, when I only used the two. Anyone have any ideas?

7. Oct 20, 2014

### ehild

You also used the asymptote at infinity where the function behaves as 50-0.2x.

ehild

8. Oct 20, 2014

### tycoon515

Being that your graph has an oblique asymptote, an improper rational function should be the kind of function that comes to mind. By the looks of the graph, as we let $x$ tend toward either $\infty$ or $-\infty$, the function gets closer and closer in value to a linear function, so we should consider an improper rational function whose numerator is $1$ greater in degree than its denominator. Also, a rational function is equal to $0$ wherever its numerator, but not its denominator, is equal to $0$. Since you have three $x$-intercepts, the numerator should be a polynomial that is the product of three factors, each one corresponding to a different $x$-intercept. A preliminary construction might look like the following:

$f(x)=k\frac{(x+40)(x-.5)(x-277.4)}{?}$

The numerator, when expanded, appears to be a polynomial whose dominant term is $x^3$. Therefore, we should consider a denominator that is a polynomial of degree $2$. Also, the graph is undefined and has a vertical asymptote at only one value of $x$: $-6$. Notice also that the graph of $f(x)$ tends toward $-\infty$ as $x$ approaches $-6$ from either side. Therefore, the denominator should be a quadratic with at least one $(x+6)$ factor. To find the other factor, observe either that there is no sign change of the function across the vertical asymptote at $x=-6$, or that for any factor other than $(x+6)$ to be present in the denominator, the graph would have to have another vertical asymptote that isn't at $x=-6$, so the denominator clearly has repeated $(x+6)$ factors. Our construction now takes the following form:

$f(x)=k\frac{(x+40)(x-.5)(x-277.4)}{(x+6)^2}=k\frac{(x+40)(\frac{2x-1}{2})(\frac{5x-1387}{5})}{(x+6)^2}=k\frac{(x+40)(2x-1)(5x-1387)}{10(x+6)^2}$

$=k\frac{(2x^2+79x-40)(5x-1387)}{10(x^2+12x+36)}=k\frac{10x^3-2379x^2-109773x+55480}{10x^2+600x+1800}$

A function of this form is EXACTLY equal to $0$ at all three of the intercepts shown on the graph provided. And, if we let $k=-\frac{1}{5}$, it has an oblique asymptote of $-\frac{1}{5}x+\frac{2379}{50}$, which has the same slope as the asymptote on the graph but NOT the same $y$-intercept. Also, the equation you came up with has the correct slope and $y$-intercept of the oblique asymptote, but it is NOT EXACTLY equal to $0$ when $x=277.4$, so one of two things must be going on here: either the $x$-intercepts displayed on the graph are slightly off (rounded), or the actual equation of the oblique asymptote is slightly different than $-\frac{1}{5}x+50$.

9. Oct 20, 2014

### TheExibo

Awesome! Thank you to everyone for the help!


# Does the intersection of two finite index subgroups have finite index?

Let $(G,*)$ be a group and $H,K$ be two subgroups of $G$ of finite index (the number of left cosets of $H$ and $K$ in $G$). Is the set $H\cap K$ also a subgroup of finite index? I feel like need that $[G\colon(H\cap K)]$ is a divisor of $[G\colon H]\cdot[G\colon K]$, but I dont't know when this holds.

Can somebody help me out?

• Step 1: Show that $H \cap K$ is a subgroup of $H$ of finite index. Step 2: Show that $[G:H] = [G:H \cap K][H\cap K : H]$. – Clive Newstead Apr 5 '12 at 22:41
• @MartQ. You don't have to show that it's a divisor. It suffices to show that $[G\colon(H\cap K)]\leq [G\colon H]\cdot[G\colon K].$ You can try to find an injection from a certain set to another. – user23211 Apr 5 '12 at 23:02
• A related question. – user23211 Apr 5 '12 at 23:06
• @CliveNewstead The formula $[G:H] = [G:H \cap K][H\cap K : H]$ seems incorrect to me. What is $[H\cap K : H]?$ Did you mean $[G:H\cap K]=[G:H][H:H\cap K]?$ – user23211 Apr 5 '12 at 23:23
• @ymar: Yes, sorry, I was having a dense moment. Unfortunately it's too late to edit my comment. – Clive Newstead Apr 7 '12 at 11:45

Proof $1$: $\quad[G:H\cap K]=[G:H][H:H\cap K]=[G:H][HK:K]\le [G:H][G:K].$

We do not assume normality on $H,K$ and therefore cannot assume $HK$ is a group. But it is a disjoint union of cosets of $K$, so the index makes sense. Also, $[H:H\cap K]=[HK:K]$ follows from the orbit-stabilizer theorem: $H$ acts transitively on $HK/K$ and the element $K$ has stabilizer $H\cap K$.

Proof $2$: Consider the diagonal action of $G$ on the product of coset spaces $G/H\times G/K$. The latter is finite so the orbit of $H\times K$ is finite, and the stabilizer of it is simply $H\cap K$. Invoke orbit-stabilizer.

• There's probably a thread in which your second equality is proved, too. I'll go digging. – Dylan Moreland Jul 14 '12 at 23:10
• – Dylan Moreland Jul 14 '12 at 23:15
• The index $[HK : K]$ is the size of the set of left cosets of $K$ in $HK$, i.e., the size of $\{gK \mid g \in HK\}$. But then $g=hk$ and so any element of $\{gK \mid g \in HK\}$ will look like an element from $H/K$. But $K$ is not a subgroup of $H$. Does this still make sense? – Al Jebr Sep 23 '18 at 21:08
• @AlJebr The notation $HK/K$ makes sense, the notation $H/K$ in general does not. Every element of $HK/K$ is a coset $hK$ for some $h\in H$. – anon Sep 24 '18 at 0:29

Let $\{H_1,\dots,H_m\}$ be the left cosets of $H$, and let $\{K_1,\dots,K_n\}$ be the left cosets of $K$. For each $x\in G$ there are unique $h(x)\in\{1,\dots,m\}$ and $k(x)\in\{1,\dots,n\}$ such that $x\in H_{h(x)}$ and $x\in K_{k(x)}$. Let $p(x)=\langle h(x),k(x)\rangle$. Note that the function $p$ takes on at most $mn$ different values. Now show:

Proposition: If $x$ and $y$ are in different left cosets of $H\cap K$, then $p(x)\ne p(y)$.

It follows immediately that $H\cap K$ can have at most $mn$ left cosets.

It may be easier to consider the contrapositive of the proposition:

If $p(x)=p(y)$, i.e., if $x$ and $y$ are in the same left coset of $H$ and the same left coset of $K$, then $x$ and $y$ are in the same left coset of $H\cap K$.

You may find it helpful to recall that $x$ and $y$ are in the same left coset of $H$ iff $x^{-1}y\in H$.

• Please, professor, could you clarify what does the $p(x)$ mean? – Danilo Gregorin Afonso Sep 4 '17 at 1:20
• @Danilo Gregorin: $p$ is a function which assumes ordered pairs of natural numbers as its values. – Shahab Sep 5 '17 at 15:32
• @Shahab Thank you very much. I had not undestood (is this correct english?) the notation, so guessed it meant something else. – Danilo Gregorin Afonso Sep 5 '17 at 19:52

Another way to see that the answer is “yes”: In this thread it is shown that any finite index subgroup of $G$ contains a subgroup which is normal and of finite index in $G$. Find such subgroups $N_1 \subset H$ and $N_2 \subset K$. Then $G/N_1 \times G/N_2$ is a finite group; do you see why this implies that $N_1 \cap N_2$ has finite index in $G$?

• I think the diagonal action on the product of coset spaces with orbit-stabilizer works fine without $H,K$'s (or subgroups of them) being normal. – anon Jul 14 '12 at 22:59
• @anon You should post that as an answer! I gave an answer to a closed version of the linked question recently, so this fact was on my mind. – Dylan Moreland Jul 14 '12 at 23:03

$H, K$ be subgroups of $G$. Any $A \in \frac{G}{H \cap K}$ can be written as $A = B \cap C$, where $B \in G/H$ and $C \in G/K$ as follows. For any $g \in G$ $$g(H\cap K) = gH \cap gK$$.

Hence, $$[G:H \cap K] \leq [G:H] [G:K]$$.

Let $C$ be the set of left cosets of $S = H \cap K$ in $G$, $C_1$ the set of left cosets of $H$ in $G$, and $C_2$ the set of left cosets of $K$ in $G$. Consider the function $f: C \to C_1 \times C_2$ defined by $f(xS) = (xH, xK)$. It is easy to check that this function is well-defined. Furthermore, this function is injective. Hence, $|S| \leq |C_1| \cdot |C_2|$, which proves that $[G: S]$ is finite.

Let $$l := [G:H], m := [G:K], h_1,...,h_m,k_1,...,k_l \in G$$ and $$G$$ be partitioned as $$G=h_1H \cup ... \cup h_lH = k_1K \cup ... \cup k_mK$$

I will find $$a_1,...a_n \in G$$ s.t. $$G$$ is partitioned as $$G=a_1(H \cap K) \cup ... \cup a_n(H \cap K)$$ which means $$n=[G:H \cap K] < \infty$$.

Let $$b_1 \in G$$. Then $$\exists i_1 \in \{1,...,m\}, j_i \in \{1,...,l\}$$ s.t. $$b_1 \in h_{i_1}H \cap k_{j_1}K=b_1H \cap b_1K = b_1(H \cap K)$$. Next, let $$b_2 \in G \ \setminus \ b_1(H \cap K)$$. Then $$b_2 \in b_2(H \cap K)$$ where $$b_2(H \cap K) \cap b_1(H \cap K) = \emptyset$$ because $$b_1H \cap b_2H = h_{i_1}H \cap h_{i_2}H = \emptyset = k_{j_1}K \cap k_{j_2}K = b_1K \cap b_2K$$

where $$h_{i_2}H=b_2H, k_{j_2}K=b_2K, i_2 \in \{1,...,m\} \ \setminus \ \{i_1\}, j_2 \in \{1,...,l\} \ \setminus \ \{j_1\}$$ This process continues at most $$lm$$ times for $$a_p=b_p, p \in \{1,2,...,n\}$$ Thus, $$n \le lm < \infty.$$

I'm not really adding something to the answers, and surely you already have what you need, but I'll leave this here just for people having the same problem who will eventually end up here.

Since both $$H$$ and $$K$$ are finite index subgroups of $$G$$, we can consider the right cosets $$G/H$$ and this is finite by hypothesis. In particular $$G$$ acts (transitively) on $$G/H$$ by right multiplication, i.e. $$G/H \times G \to G, \qquad (Hx, g) \mapsto Hxg.$$ If we now restrict the action to $$K$$, and consider the $$K$$-orbit of $$H$$, call it $$\mathcal{O}_K(H)$$, then it must be finite since it's contained in $$G/H$$. But now by what Isaacs calls the Fundamental Counting Principle (see "Finite Group Theory", theorem 1.4) we exactly know the following equality $$|\mathcal{O}_K(H)| = [K : \text{Stab}_K(H)],$$ where $$\text{Stab}_K(H)$$ is the stabiliser of $$H$$ under the $$K$$-action. Notice that $$\text{Stab}_K(H) = K \cap \text{Stab}_G(H)$$, and what's the stabiliser of $$H$$ for the $$G$$-action? But of course $$H$$ itself, so $$\text{Stab}_K(H) = K \cap H$$, and finally we conclude that $$[K : K \cap H]$$ is finite. Hence $$[G : H \cap K] = [G : K][K : K \cap H]$$ must be finite as well.


# How to graph the equation: $y=\frac {x-2}{x+1}$?

the title says it all. I'm pretty sure this is a hyperbola, but is there an alternative way of doing this besides a table of values?

"Graph the equation $y=\frac {x-2}{x+1}$"

I know that $x$ cannot equal $-1$ but I'm not sure how to carry on from there.

Any help would be appreciated. If you could provide a step by step explanation, and a picture of the graph itself, that would be great :D!

• Do you mean to show it? – Ofir Attia May 20 '13 at 8:36
• I just need to graph the equation, but an explanation on how to do it would be nice too :D – missiledragon May 20 '13 at 8:36
• $y$ can indeed equal 0, i.e, when $x = 2$. ;) – user49685 May 20 '13 at 8:38
• OH, I didn't realise that :P. Thanks @user49685 – missiledragon May 20 '13 at 8:40

When you want draw the graph of the function, you need to check a couple of things:

1. Check if there are any points in the domain where the function is undefined, you know there's something happening there. For instance here you saw that when $x=-1$ it is undefined.

2. Then, $y$ can be equal to $0$, there's no problem with that, and you should find all the $x$ such that $y=0$. In this case, when $x=2, y=0$. Thus you know that the function goes through the point $(2,0)$.

3. Now you have all the points where there's something special happening, you can check where the function is positive, and where it is negative. You need to look at all the intervals between the special points. In your case : $(-\infty,-1),\,(-1,2),\,(2,\infty)$. Take $(-\infty,-1)$ for example. On this interval, you know that the numerator is always negative and the denominator is always negative as well, so the value for $y\in(-\infty,-1)$ will always be positive. Do that for all the intervals.

4. Finally, you need to check some limits to see what happens at those special points and also when you tend to infinity (in your case only $-1$ is a special point because you already now that the value of your function when $x=2$ is $0$). So you should check :

$$\lim_{x\rightarrow\infty}f(x),\; \lim_{x\rightarrow-\infty}f(x),\; \lim_{x\rightarrow-1^+}f(x),\; \lim_{x\rightarrow-1^-}f(x),\;$$

That's it, you have everything you need to graph your function! (If you want to go into further details you can also check the derivative for extremas and the second derivative for inflexion points. This means that you have to compute the derivatives and check for which values of $x$ they are equal to $0$).

• Thanks! that helped me a lot :D! – missiledragon May 20 '13 at 9:06

Rewrite as $y=1-3/(x+1)$, or $(y-1)= -3/(x+1)$

This is simply the regular recriprocal equation, with the axies rescaled. The standard $Y$ axis, representing $X=0$, is now marked $x=-1$. That is, the x-numbers are moved one to the right. The old X axis of 0,1,2,3 become -1, 0, 1, 2...

The old $X$ axis, or $Y=0$ becomes $y=1$, and we have to rescale by a factor of 3. So the old Y coordinates 0, 1, 2, 3... become 1, -2, -5, -8...

And that's it.

Full Function investigation, check of asymptotic (infinite and non setting points), extreme points and inflection points. nice tool to work with is desmos:
$y=(x−2)/(x+1)$ - Click to view

In addition : Investigating Functions - Click to view

hope its helped you.

Just to add to what Oliver and Wendy have said, To Draw any function:
1.)Check Domain and critical points.
2.)Check intercepts.By putting $x=0 ,y=0$
3.)Check Asymptodes around infinity and critical points by taking $$\lim_{x\rightarrow\infty}f(x),\; \lim_{x\rightarrow-\infty}f(x),\; \lim_{x\rightarrow-1^+}f(x),\; \lim_{x\rightarrow-1^-}f(x),\;$$ 4.)Check Symmetry about axis by putting $x = -x$ and $y=-y$
5.)Intervals of increase/ decrease by equating first derrivative to zero.
If you know all this information you can draw any graph manually.

Hope this helps :)


# Big O Notation “is element of” or “is equal”

People are always having trouble with "big $O$" notation when it comes to how to write it down in a mathematically correct way.

Example: you have two functions $n\mapsto f(n) = n^3$ and $n\mapsto g(n) = n^2$

Obviously $f$ is asymptotically faster than $g$. Is it $f(n) = O (g(n))$ or is it $f(n) \in O(g(n))$?

My prof says that the first one is wrong but is a very common practice, therefore it is used very offten in books. Although the second one is the right one.

Why is that so?

• Formally, neither are. If you define $O(\cdot)$ as a set of functions (for asymptotics w.r.t. a given point), say $O(g) = \{ f : \exists N,C>0,\ \forall n\geq N\ f(n) \leq C\cdot g(n)\}$, then you would have to write $f\in O(g)$. $f(n)$ is not a function, it's a number. But most people are quite happy with the (slight) abuse of notation: $f(n) = O(g(n))$ for $f\in O(g)$, because there is no ambiguity anyway. – Clement C. Dec 20 '16 at 15:13
• Ah, because f(n) would be a specific value when f is meant generally, is that right? – Blnpwr Dec 20 '16 at 15:14
• Indeed. The same way that, technically, $f(x)$ is the value of the function $f$ when evaluated at the point $x$, not the function $f$ itself. – Clement C. Dec 20 '16 at 15:15
• For this particular case, you have in fact $g(n) = o(f(n))$ or $f(n) = \omega(g(n))$ because $\lim_n \frac{f}{g} = \infty$ – Alex Dec 20 '16 at 15:48
• For your example functions it's the other way around. $g(n)=O(f(n))$. I'm assuming we're talking about $n\to\infty$, which is hinted by the use of the letter $n$. – Meni Rosenfeld Dec 21 '16 at 9:48

I really like Wikipedia's note on this:

The statement “$f(x)$ is $O(g(x))$” […] is usually written as $f(x) = O(g(x))$. Some consider this to be an abuse of notation, since the use of the equals sign could be misleading as it suggests a symmetry that this statement does not have. As de Bruijn says, $O(x) = O(x^2)$ is true but $O(x^2) = O(x)$ is not. Knuth describes such statements as “one-way equalities”, since if the sides could be reversed, “we could deduce ridiculous things like $n = n^2$ from the identities $n = O(n^2)$ and $n^2 = O(n^2)$.”

For these reasons, it would be more precise to use set notation and write $f(x) \in O(g(x))$, thinking of $O(g(x))$ as the class of all functions $h(x)$ such that $|h(x)| \leq C|g(x)|$ for some constant $C$. However, the use of the equals sign is customary. Knuth pointed out that “mathematicians customarily use the $=$ sign as they use the word ‘is’ in English: Aristotle is a man, but a man isn't necessarily Aristotle.”

• Gotta love an answer backed up by Knuth, even if it is by a quote inside a quote. – Jasper Dec 20 '16 at 17:30
• This still isn't a complete answer. Using $f(x)$, $g(x)$, $h(x)$, $n$, $n^2$ etc. to mean functions is also abuse of notation. – JiK Dec 21 '16 at 15:07
• @JiK you're right, but it is a common abuse of notation, and in my opinion the crux of the question regards whether to use $=$ or $\in$, not whether or not to be completely formal every time you talk about a function – TomGrubb Dec 21 '16 at 15:15
• I don't agree with the statement would be more precise to use set notation. It implora that messure theory lenguage almost every point, cotient spaces, etc. are wrong lenguage. I think viewing big O / little o notation in this way is a bery powerful tool and os pretty precise. – Ernesto Iglesias Apr 23 '18 at 16:40

Using $\in$ is set-theoretically correct but inconvenient. For example, $$\sin x = x - \frac{x^3}{3} + \mathrm O(x^5)$$ In this case $\mathrm O$ should be interpreted as there exists an $\mathrm O(x^5)$ function to make this equality valid. The $=$ notation also allows asymptotic notation to appear on both sides and do arithmetic: $$e^x + \mathrm O(x) = \mathrm O(e^x)$$ In this case, for every $\mathrm O(x)$ function on the left, there exists an $\mathrm O(e^x)$ function on the right to make this an equality.

Warning: In this case the two sides of $=$ cannot be swapped carelessly. e.g. $\mathrm O(x) = \mathrm O(e^x)$ but $\mathrm O(e^x) \neq \mathrm O(x)$.

• "but inconvenient" Indeed! I would go as far as saying that using it almost completely defeats the very purpose of the notation, rendering it pointless to introduce it in the first place. – quid Dec 20 '16 at 20:02
• For what it's worth, there is a way to reconcile this. The way I think about it is that using $\mathcal O(x^5)$ here changes the abuse of notation somewhat: we're now dealing with sets of functions, so the function $x$ becomes the singleton set $\{ x \mapsto x \}$ (and likewise for $x^3 / 3$), the addition operator becomes setwise product-addition $A \oplus B = \{ x \mapsto f(x) + g(x) : f \in A, g \in B \}$, $\mathcal O(x^5)$ really is a set, and the whole statement should be read as $(x \mapsto \sin x) \in {\text{RHS}}$. Perhaps slightly inconvenient, but I don't see anything unsound. – wchargin Dec 21 '16 at 8:00
• This reminds me of equality of multifunctions, like if we say $\cos^{-1} 1 = 2k\pi$ – GFauxPas Dec 21 '16 at 14:24
• @wchargin Using $O(x^5)$ does not automatically mean we are dealing with sets of functions. All that the writer means is that they're dealing with a specific quantity which satisfies the definition of being $O(x^5)$. The purpose of notation is to communicate what a mathematician is thinking. Introducing sets of functions—frankly, no one cares about the set of all functions that are $O(x^5)$ and it's an uninteresting set; we only care that the particular function we're working with is $O(x^5)$—is backwards, saying we should change our thinking to fit (someone's narrow idea of) $=$ notation. – ShreevatsaR Dec 21 '16 at 20:42
• @SheevastaR ¡¡¡Of course we care for the set $O(g)$!!! if you don't is a problem you have. Read the answer you are commenting and you will notice how important it is. – Ernesto Iglesias Apr 23 '18 at 16:49

$O(g(x))$ is a class of functions - think of it as a "property" functions can have. By the literal interpretation of the equals sign, "$f(x) = O(g(x))$" should be interpreted as "$f$ is literally equal to a certain class of functions." But functions and classes of functions are different sorts of things - even if this was what we meant to say, it's like saying that one particular apple is equal to a basket of apples. But what we mean when we say "$f(x) = O(g(x))$" is that $f$ belongs to the class of functions $O(g(x))$ - so, $f(x) \in O(g(x))$.

The reason we use $=$ instead of $\in$ is because, given the particular uses of big-$O$ notation (and little-$o$ notation, if you're familiar with that) $=$ is massively more convenient. We say things like $x^3 + O(x) = O(x^3)$, for example; we don't mean that $O(x)$ is an object that can actually be added to $x^3$, or that when that addition is done we actually get the class of functions $O(x^3)$, we just mean that for any function $f \in O(x)$, the function $x^3 + f(x)$ is a member of $O(x^3)$. But if I wanted to write that out in more standard notation, I'd have to say something like $\{x^3 + f(x) \mid f(x) \in O(x)\} \subseteq O(x^3)$. This is inconvenient to write and difficult to read, so we prefer the "slicker" notation $x^3 + O(x) = O(x^3)$.

However, I'm not sure I would say that sentences like $f(x) = O(g(x))$ are wrong. By convention, they're perfectly right - it's just that when an expression includes $O$ (or $o$), $=$ does not mean what it usually means. That's okay.

• I think there must be someone that had invented a convenient way to just say $x^3 + O(x) = O(x^3)$, and at the same time, the symbols are entirely rigorous. – Eric Dec 20 '16 at 16:29
• @Eric Well, $x^3 + O(x) = O(x^3)$ is an entirely rigorous and convenient way to say that. It just doesn't look rigorous if you don't know the definition of "$=$" in the context of $O$. Since everyone in the field understands this notation, no one's particularly motivated to find a new way of writing it that won't confuse outsiders. – Reese Dec 20 '16 at 16:34
• @Reese If you treat $O(x)$ as a set, then $x^3 + O(x) \subseteq O(x^3)$ is rigorous without domain-specific $=$ sign. – heinrich5991 Dec 21 '16 at 1:02
• @heinrich5991 But that requires domain-specific $+$. – Reese Dec 21 '16 at 1:34
• @heinrich5991 Oh, absolutely, it's common. But so is using $=$ for $O$ and $o$. My point is that it's still not transparent to first learners. – Reese Dec 22 '16 at 17:29

I'll try to clarify wchargin comment.

If you are familiar with Group theory you should not find the $=$ sign awkward. Class of equivalence and quotient spaces usually use the sum notation for doing that.

e.g., $\mathbb{R}/\mathbb{Z}$ elements are usually written as $a+\mathbb{Z}$.

The problem is that big $O$ notation makes implicit use of that, i.e. in the class of $$O(g)=\lbrace f:\exists M,C>0, \forall x_0\geq M: f(x_0)\leq Cg(x_0) \rbrace,$$ as exposed by Clement C.

In this way, $$f=O(g)$$ Means $$f+O(g)=0+O(g)$$ And that is to say than $f\in O(g)$.

Regarding $=$ being more useful/convenient than $\in$ because of allowing things like $x^3 + O(x) = O(x^3)$, it seems like for that you could use $x^3 + O(x) \subseteq O(x^3)$. This seems more precise to me, because it would mean "every element of the set $x^3 + O(x)$ is an element of the set $O(x^3)$." which is, I think, exactly what one wants to say. (where $g+A:=\{g+f:f\in A\}$ and when $g$ is a function and $A$ is a set of functions, and similarly for similar situations.)

Making the statement with subset is mentioned by Reese's answer, but I think that defining the sum of a function and a set of functions, and also the sum of sets of functions, in the straightforward way removes the inconveniences of saying things like $\{x^3+f(x)∣f∈O(x)\}\subseteq O(x^3)$. (Defining things similarly with multiplication, and application of functions in general.)

When things are defined this way, it should become as simple as using $\in$ instead of $=$ when there is a single function on the left, and using $\subseteq$ instead of $=$ when there is a set of functions on the left, and it seems more precise.

• No, set-theoretic inclusion is not "exactly what one wants to say". The purpose of notation is to aid communication, and it is meant to fit our way of thinking. It is misguided to insist that one should change one's (very productive) way of thinking just to fit some notation or formalism. When an analyst writes $\sin x = x - \frac{x^3}{3} + \mathrm O(x^5)$ they are literally thinking that $\sin x$ is equal to $x$ minus $\frac{x^3}{3}$ plus some function that is $\mathrm O(x^5)$ — they are thinking of equality, not set inclusion, so the notation reflects that. – ShreevatsaR Dec 21 '16 at 18:54
• Ok, I should probably defer to you here. But isn't saying that $sin(x)$ is equal to $x - \frac{x^3}{3}$ plus some function that is $O(x^5)$ the same thing as saying that $sin(x)$ is an element of the set of functions which are of the form $x - \frac{x^3}{3}$ plus some function that is $O(x^5)$? – drocta Dec 21 '16 at 20:31
• Well, is saying "the sky is blue" the same thing as saying "the sky is an element of the set of things that are blue"? In some sense yes, and in another sense no. (It may be logically equivalent, but the speaker may not have even considered, or care about, the set of all things that are blue.) One can also say "the sky satisfies the property of being blue", or even "blueness inheres in the sky". There can be many ways to formalize a statement, and the most fruitful ones are often the one that closest match what the human is thinking, rather than what fits cleanest with other formalisms. – ShreevatsaR Dec 21 '16 at 20:45
• (sorry, I hit enter by mistake. First time using SE comments.) Your point about the purpose being for communication seems like something I failed to consider. But $\in$ being like "has property" and $\subseteq$ being like "all things with property on the left have the property on the right$, seem like they capture the$\forall$or$\exists$in what is being said with "there exists a function of the form$x-x^3/3+O(x^5)$which is$sine(x)$, and "for all functions of the form <blah>, there is a function of the form <bleh> such that they are equal". But maybe it's not nice enough to justify sets? – drocta Dec 21 '16 at 20:57 Here is my five pennies worth: If we see a statement of the form $$f(x)=g(x)+O\bigl(p(x)\bigr)\qquad(x\to\xi)$$ then for each$x$the exact value of the term$O\bigl(p(x)\bigr)$is defined by this very equation: it is$:=f(x)-g(x)$. In addition we are told that there is a constant$C>0$such that for all$x$in a suitable punctured neighborhood of$\xi$this difference is$\leq C\bigl|p(x)\bigr|\$.


# Finding the sin inverse

Well, so I was finding the value of $\sin 15^{\circ}$. I used the identity $2 \sin x \cos x = \sin 2x$. So solving the quadratic equation and picking the right values simply gave me:

$$\sin 15^{\circ} = \frac {\sqrt {2 - \sqrt {3}}} {2}$$

That was pretty much simple. But I thought what if I was given the value instead, and I was asked to find the angle, whose sine would give me that value.

In short, my question is how to evaluate:

$$\sin^{-1} \left( \frac {\sqrt {2 - \sqrt {3}}} {2} \right)$$

I want to solve it thinking that I do not know the value of $\sin 15$, as if I have been just provided with the problem, and I have no idea what the answer might be. I tried to proceed by converting the $\sin$ to $\tan$ but that seemed to be of no use. Can anybody help?

• $$15=45-30=60-45$$ – lab bhattacharjee Feb 23 '18 at 18:44
• You are doing that because you know that the answer is 15. How is someone supposed to know that just on seeing the problem? – Manish Kundu Feb 23 '18 at 18:48
• Please find my answer – lab bhattacharjee Feb 24 '18 at 10:46

In general it's hard, if not impossible. In this case, it turns out to be doable.

The main idea is that if $\theta = \sin^{-1}\left( \frac{\sqrt{2 - \sqrt 3}}{2} \right)$, then $\sin \theta = \frac{\sqrt{2 - \sqrt 3}}{2}$, and we have an equation we can (hopefully) solve for $\theta$.

Squaring, we get $$\sin^2 \theta = \frac{2 - \sqrt{3}}{4},$$ and multiplying by 4, $$4\sin^2 \theta = 2 - \sqrt{3}.$$

My thoughts are that maybe if I get $\sqrt{3}$ by itself and square, I'll get a nice equation that's polynomial in $\sin \theta$ with integer coefficients. So, we get $\sqrt{3}$ by itself: $$\sqrt{3} = 2 - 4 \sin^2 \theta.$$ But that's interesting: I can factor 2 out... $$\sqrt{3} = 2(1 - 2 \sin^2 \theta),$$

and that $1 - 2 \sin^2 \theta$ looks pretty good, because that's the output from a double angle identity, namely $\cos(2 \theta)$. Now, $$\sqrt{3} = 2 \cos(2 \theta),$$ so $\cos (2\theta) = \frac{\sqrt{3}}{2}$.

This happens when $2 \theta = 30^\circ$ or $2\theta = 330^\circ$.

So, we get one possibility that $\theta = 15^\circ$ (since we've solved an equation by squaring, though, we should make sure it's not extraneous!).

And of course, once we know one angle $\theta$ with a particular sine, we can use symmetry and the unit circle to find all other such angles (that would be $180^\circ - \theta = 180^\circ - 15^\circ = 165^\circ$ in this case, and everything coterminal with the two between $0^\circ$ and $360^\circ$). If we just want the inverse sine of something positive, it's an angle in quadrant I.

• Sidenote, running off to class now, won't be able to respond for a few hours (if it matters) – pjs36 Feb 23 '18 at 19:07
• That was really amazing. Didn't expect it to be solved so simply. – Manish Kundu Feb 23 '18 at 19:23
• @ManishKundu Thanks! I was surprised too; it's a really nice question, and I'm happy you thought to ask it. – pjs36 Feb 24 '18 at 2:26

$$2-\sqrt3=\dfrac{(\sqrt3-1)^2}2$$

$$\implies\dfrac{\sqrt{2-\sqrt3}}2=\dfrac{\sqrt3-1}{2\sqrt2}$$ as $\sqrt3-1>0$

$$=\sin60^\circ\cos45^\circ-\sin45^\circ\cos60^\circ$$

$$=\sin(60-45)^\circ$$

Now we know the principal value of $\sin^{-1}x$ lies in $\in[-90^\circ,90^\circ]$ for real $x$

• You made it simple. That was nice. – Manish Kundu Feb 24 '18 at 15:41

+1 to pjs36's answer, and they did bring up a good point that in this case it was possible to solve. But what if we weren't able to use the double angle identity? Perhaps another way to go would be the use of MacLaurin series.

Given a problem $\sin^{-1}{A} =\theta$, where $\theta$ is in radians and $A \in [-1,1]$, we can first take the $\sin$ of both sides to get $$\sin{\theta} = A$$

The MacLaurin series for $\sin{\theta}$ is $$\sin\theta=\sum_{n=0}^\infty \frac{(-1)^n \theta^{2n+1}}{(2n+1)!} = \theta - \frac{\theta^3}{3!}+\frac{\theta^5}{5!} - \dots =A$$

The farther out you extend the polynomial, the better the estimate is going to be when you solve for $\theta$. However it will be only an estimate when you do it this way, and you would probably need to use a calculator or WolframAlpha or something to solve it as well, which probably defeats the purpose of avoiding just plugging in $\sin^{-1} A$ into the calculator in the first place. This is merely another way you could look at the problem, if you wanted to avoid using inverse trig.

• So I need to solve for $\theta$. Wait, so do I need to use Newton Raphson? – Manish Kundu Feb 23 '18 at 19:49
• Correct, we want to solve for $\theta$. However, this would be equivalent to solving the roots for a large polynomial. Newton Raphson could work, as the derivative of a polynomial is simple to find, but it may take alot of time still to do by hand – WaveX Feb 23 '18 at 19:52
• We would also have the case that we would be estimating an estimate. – WaveX Feb 23 '18 at 19:59
• I see now why solving such problems can be so hard in general. – Manish Kundu Feb 23 '18 at 20:00

For a shortcut evaluation of $x$ such that $\sin^{-1}(x) =k$, assuming that $x$ is "small", you could use the simplest Padé approximant $$\sin^{-1}(x)\approx \frac{x}{1-\frac{x^2}{6}}$$ leading to $$x\approx \frac{\sqrt{6 k^2+9}-3}{k}$$

For your case, where $k= \frac {\sqrt {2 - \sqrt {3}}} {2}$, this would give $$x\approx \sqrt{2+\sqrt{3}} \left(\sqrt{48-6 \sqrt{3}}-6\right)\approx 0.255992$$ in radians (that is to say $\approx 14.6673$ in degrees).

• That was helpful. Thanks. – Manish Kundu Feb 24 '18 at 7:10

Not an answer, just an explicit formula for the $$\arcsin$$ function.

First recall that $$\sin x=\frac{e^{2ix}-1}{2ie^{ix}}$$ So if we want the function $$y=\arcsin x$$ we must solve the following equation for $$y$$: $$x=\frac{e^{2iy}-1}{2ie^{iy}}$$ if we set $$u=e^{iy}$$ this becomes much easier: $$x=\frac{u^2-1}{2iu}$$ $$u^2-1=2iux$$ $$u^2-2iux-1=0$$ The use of the quadratic formula then gives $$u=ix+\sqrt{1-x^2}$$ And using $$\log$$ to denote the complex natural logarithm: $$iy=\log\left(ix+\sqrt{1-x^2}\right)$$ $$\arcsin x=-i\log\left(ix+\sqrt{1-x^2}\right)$$


# Angle needed for object A to intercept with object B

Object B is 15 degrees East of North at a distance of 20km/h. Object B is moving at an average speed of 30km/h in the direction 40 degrees East of North. If object A is capable of moving at 100km/h, at what angle does it need to move to intercept object B?

I've tried calculating it as vectors, but got stuck. In the end I just tried using good old geomatry, where $$\frac{\sin A}{a} = \frac{\sin B}{b}$$ So I called the angle I'm searching for A, and the angle from the line between A and B, and the direction of object B, as angle B. 'a' and 'b' are the average speeds of object a and b. $$\frac{\sin A}{30 \times t} = \frac{\sin(155°)}{100\times t}$$ $$A = arcsin\left(\frac{30\times t\times \sin(155°)}{100\times t} \right) = 7.284°$$

So the final angle should be $15°+7.284°=\mathbf{22.284°}$

Is this correct? Isn't there a better way of doing this, like using vector products?

PS. I've found a very similar question here, but didn't quite understand how i.e. $\alpha + \beta = 40°$, because my calculations show $=35°$. And trying to read through comments and answers, confuses me more due to updates being made inconsistent to comments about the updates done. And due to me not having enough reputation, I'm not able to ask counter-questions.

BONUS Question: I have been searching high and low on how to make a simple graph of my problem. Like this here question, with a graph of the problem hosted at stack.imgur.com which I guess was generated by LaTeX on stackexchange.com, right?

-
gnuplot makes nice graphs, as does MATLAB, but gnuplot is free. – Pranav Hosangadi Sep 27 '13 at 3:29
Thanks @PranavHosangadi! I didn't realize I had to make them and them upload the photo. I thought MathJax had a plugin to make LaTeX plots in posts. This makes more sense now! – nikc Sep 27 '13 at 7:01
I wouldn't use Gnuplot or Matlab for this; They are good for plotting data, but would basically require you to solve the problem in order to plot something like this. I would use Inkscape for this kind of a drawing. – Colin McFaul Sep 27 '13 at 15:06

Simple way to solve: Use distances travelled. So B has already travelled 20km at 15° East of North. That is $20 \times \sin(15°)$km northwards and $20 \times \cos(15°)$ eastwards.

Now consider A intercepts B at a time $t$. The equations for the distances travelled by B are:

$Y_B = 20 \times \sin(15°) + 30 \times t \times \sin(40°)$ in the Y direction and
$X_B = 20 \times \cos(15°) + 30 \times t \times \cos(40°)$ in the X direction.

For A, the distances travelled are:
$Y_A = 100 \times t \times \sin(\alpha)$ in the Y direction and
$X_A = 100 \times t \times \cos(\alpha)$ in the X direction.

For the two to intercept, $X_A = X_B$ and $Y_A = Y_B$

You now have 2 equations, 2 variables, and a whole lot of fun solving them simultaneously.

-

There is a technique based on vectors that solves the problem, and lets you know if there is no solution, one solution or two solutions.

(All my angles are measured counter-clockwise from the positive X-axis)

Assume the pursuer travels at 100 km/hr at an angle $\beta$. We're going to try to divide this velocity into two non-perpendicular components

One matches the velocity of the target exactly. So in the problem above, assume that one part of A's velocity is $30\angle 50°$. It wouldn't matter if A's total velocity were less than 30.

So, we have (in theory) completely cancelled out B's attempt to escape. Now, how does A reach B? A must travel at some velocity (say X) along the original direction from A to B. (Since we've cancelled out any of B's movement) So, in this problem, the other part of A's velocity is $X \angle 75°$

So, finally, we're reduced to solving:$$30\angle 50°+X \angle 75°=100\angle\beta$$One way is to expand each vector into X and Y components and set up two equations. If you eliminate $\beta$ you get a quadratic in X. If there are any real positive values of X, you can then substitute and solve for $\beta$

Expanding the equation above for components in the X and Y direction: $$30\cos(50)+X\cos(75)=100\cos(\beta)$$ $$30\sin(50)+X\sin(75)=100\sin(\beta)$$The only unknowns are the size of X (or if it even exists!) and the angle $\beta$ If you square both sides of both equations and then add the two LHS and two RHS, some trig manipulations will eliminate $\beta$ and lead to the quadratic in X.
Thanks for your answer, but I'm still not quite sure how to use your method? I'm still learning how to solve equations using polar coordinates and I don't know how to isolate the equation above. Plus, I'm trying to find angle $\beta$, but how would I do that if there are two unknown factors in the equation? – nikc Sep 27 '13 at 8:18