# How can the set of the rational numbers be countable if there is no

1. Jan 29, 2013

### student34

rational number to count the next rational number from any rational number?

We can count the next natural number, but we can't count the next rational numer.

2. Jan 29, 2013

### CompuChip

The definition of countable means, that you can assign a natural number to each of them.
Well, here's one way to count them:

3. Jan 29, 2013

### NemoReally

Starting with, say, 1/1, write down all the rational numbers in whichever order you like. Label the starting number as Rational Number 1. As you generate the next rational number, write it down (as a / b) and then write the next natural number down next to it. Continue until you run out of either rational numbers or natural numbers ...

Actually, you won't run out of either. They are in a one-to-one correspondence. The difference from the normal way of thinking about finite counting lies in the fact that you will never run out of natural numbers. This concept underlies many mathematical ideas but it may hurt your brain until you get the idea.

Interestingly, you can't do the same with real numbers. Look up Cantor's diagonal method.

4. Jan 29, 2013

### student34

If we really can assign a natural number to each of them, can you assign a natural number to the very next rational number from the number 1?

5. Jan 29, 2013

### Fredrik

Staff Emeritus
Yes. This is what CompuChip's picture is showing. The first 11 positive rational numbers (when they are ordered as in that picture) are

1. 1
2. 1/2
3. 2
4. 3
5. 1/3 (We're skipping 2/2, since it's equal to 1, which is already on the list).
6. 1/4
7. 2/3
8. 3/2
9. 4
10. 5
11. 1/5 (We're skipping 4/2, 3/3 and 2/4, since they are equal to numbers that are already on the list).

The only problem with that picture is that it only deals with positive rational numbers. You can however easily imagine a similar picture that includes the negative ones.

6. Jan 29, 2013

### NemoReally

In principle, yes. You just keep extending the table of rational numbers, labelling each one with a natural number, until you reach the "very next" rational number from 1. The trick lies in recognizing that (physical limitations aside), there is always another rational number between the "very next" rational number and 1, so you'll never actually do it, but if you keep trying you will always be able to assign a natural number to each one.

eg, halving the difference ...

label '1' as number 1 in your list. Start with 3/2 (1+1/2) and label it '2'. Then the "next" rational number (number '3') will be 5/4 ((1+3/2)/2), the next 9/8 (number '4') and so on. You can keep dividing by 2 and incrementing the label indefinitely - you will never run out of natural numbers or rational numbers that keep getting closer to 1

https://www.physicsforums.com/attachment.php?attachmentid=55160&stc=1&d=1359454296

https://www.physicsforums.com/attachment.php?attachmentid=55159&stc=1&d=1359453948

Last edited: Jan 29, 2013
7. Jan 29, 2013

### CompuChip

Of course, if you want to be very precise: there are duplicates in the tabulation of the rationals (for example, it contains 1/2, 2/4, 3/6, 34672/69344, etc). So technically you are not really making a one-to-one mapping, but you are overcounting.

In other words, you are proving that the cardinality of the rationals is at most the same as that of the natural numbers. However, since we clearly also have at least as many rationals as natural numbers (the natural numbers are precisely the first column of the grid), you can convince yourself that the cardinalities are equal.

8. Jan 29, 2013

### cragar

another way to map the rationals to the naturals is take the
positive rationals of the form $\frac{p}{q}$ and map them to
$2^p3^q$ and then map the negative rationals to
$5^{|-p|}7^{|-q|}$ and then map zero to some other prime.
In fact we are mapping all the rationals to a proper subset of the naturals.

9. Jan 29, 2013

### CompuChip

I also wanted to get back to your remark about "the next" rational number from 1... note that the rationals are dense in the real numbers. That implies that there is a rational number between any two given rationals, and therefore it is not possible to write them down in "ascending" order (i.e. write down a sequence an such that every rational is in the sequence and 0 = a0 < a1 < a2 < ...).

This may be flaw in your way of picturing the rationals that makes their countability, perhaps, a bit counter-intuitive.


# Odd prime $p$ implies positive divisors of $2p$ are $1,2,p,$ and $2p$

$$1,2,p,$$ and $$2p$$ are indeed divisors of $$2p$$. I want to show these are the only positive divisors. Is there a more elegant or concise way to prove this besides the proof I have below?

Suppose that positive $$a \in \left([3,2p-1] \cap \mathbb{N}\right) \setminus\{p\}$$ divides $$2p$$. So $$ak$$=$$2p$$ for $$k \in \mathbb{Z}$$, and clearly $$2 \leq k \leq p$$. Since $$ak=2p$$ is even, at least one of $$a$$ or $$k$$ must be even.

If $$k$$ is even, then $$a\frac{k}{2}=aj=p$$ for integer $$1 \leq j \leq \frac{p}{2}, so $$j | p$$, so $$j=1$$, so $$a=p$$ which is a contradiction.

Similarly, if $$a$$ is even then $$k | p$$, so $$k=p$$. But then $$a=2$$.

So there are no other positive divisors besides $$1,2,p$$, and $$2p$$.

The motivation for this is to show that if a group $$G$$ has order $$2p$$ for odd prime $$p$$, then nonabelian $$G$$ is isomorphic to $$D_{2p}$$. The proof begins with "the possible orders for nonidentity elements of $$G$$ are $$2,p,$$ and $$2p$$," which I am trying to prove with Lagrange's Theorem. If there is an alternative way to justify this statement using group theory, then I would appreciate seeing that as well.

• Use the Fundamental Theorem of Arithmetic. – Shaun May 26 at 16:36
• @Shaun clearly $2 \cdot p$ is a prime factorization of $2p$, and the prime factorization is unique. How does this show there cannot be nonprime divisors of $2p$ besides itself and $1$? – jskattt797 May 26 at 16:41
• Because $2p$ is squarefree (and there is only two primes). – Shaun May 26 at 16:42
• It's enough to state that the only two prime factors of $2p$ are $2,p$ so (by fundamental th) the only factors are combinations of $2$ and $p$ so they are $1,2,p$ and $2p$. Honestly that sentence is probably too long and saying too much.... I think that if you simply state the only factors of $2p$ are $1,2,p$ and $2p$ that ought to be utterly self-evident and obvious. If anyone asks way state its immediate but the fundamental th. – fleablood May 26 at 20:28

If $$p$$ is an odd prime number, then by the fundamental theorem of number theory, $$2\times p$$ is the unique primary decomposition of $$2p$$. Once you express a positive integer $$n$$ as it's unique primary decomposition, say $$p_1^{a_1}\dots p_k^{a_k}$$, then all the positive factors will be of the form $$p_1^{b_1}\dots p_k^{b_k}$$ where $$0\leq b_i\leq a_i$$ for each $$i$$. With this observation you should be able to answer your question.

• And we have $2p=2^1p^1$, so the set of all positive divisors is $$\{ 2^{a_1}p^{a_2} \mid a_1,a_2 \in \{0,1\} \}=\{1,2,p,2p \}$$. – jskattt797 May 26 at 17:17

By the fundamental theorem of Arithmetic the only prime factors of $$2p$$ are $$2$$ and $$p$$ and so every factor must be a combination of $$2$$ and $$p$$ of which $$1,2,p$$ and $$2p$$ are the only options.

That is more than sufficient and more than anyone can reasonable expect to require proof.

....

But if you want to smash an ant with a sledgehammer:

The fundamental Theorem of Arithmetic say each number has a unique prime factorization of $$n = \prod p_i^{a_i}$$. So if any factor $$m; m|n$$ can only have $$\{p_i\}$$ as prime factors and only to the powers less than or equal to $$a_i$$.

So all if $$m|n$$ then $$m$$ must be of the for $$\prod p_i^{b_i}$$ where $$0\le b_i \le a_i$$ and so there are $$\prod (a_i+1)$$ such factors.

So the factors of $$2p$$ are all of the form $$2^b p^c$$ where $$b = 0,1$$ and $$p = 0,1$$. There are four of these numbers and they are $$2^0p^0 =1; 2^1p^0 = 2; 2^0p^1 = p;$$ and $$2^1p^1 = 2p$$.

That's it.

That is a result that it is reasonable to expect every reader is either familiar with or can justify on their own.

....

But frankly it is enough to say:

"The only factor of $$2p$$ are $$1,2,p$$ and $$2p$$" and assume that is completely self-evident.

And it is.

• Yes, $m|n$ iff $m=\prod p_i^{b_i}$. And each factor $\prod p_i^{b_i}$ maps bijectively to a tuple $(b_1, \dots , b_k)$ for $0 \leq b_i \leq a_i$. So there are exactly $\prod(a_i + 1)$ factors. So an alternative proof is to notice that for $2p = 2^1 p^1$ there are exactly $2 \cdot 2=4$ factors, so there can be no others besides $1,2,p,$ and $2p$. – jskattt797 May 27 at 3:19

By the Fundamental Theorem of Arithmetic, the only primes that divide $$2p$$ are $$2$$ and $$p$$. Note that $$2p$$ is squarefree and has only two prime factors.

• Nice, so suppose there is another nonprime positive divisor $j=p_1 p_2 \dots \neq 1$ (at least two primes in $j$'s factorization because $j > 1$ is not prime) such that $j \neq 2p$, then $jk=2p$ for integer $k$. Notice $k \neq 1$, so $k > 1$ and $k = p_3 \dots$. But then $2p$ has at least 3 primes in its factorization. Why do we need squarefree? – jskattt797 May 26 at 16:58
• To rule out $2^2=4$ and $p^2$. That's all. – Shaun May 26 at 17:13
• Your argument seems valid though, @jskattt797. – Shaun May 26 at 17:14

The other answerers are of course right to point you to the Fundamental Theorem of Arithmetic as it will come in handy at many stages of your life. Still I wanted to say two things about your own proof.

1) It is correct and quite elegant. As one of the other answerers points out: using the Fundamental Theorem is a bit of a sledgehammer.

2) The key step in your proof is the lemma 'since $$ak$$ is even at least one of $$a, k$$ is even'. I wanted to point out to you that this has a very nice generalization, called Euclid's lemma. It states:

Let $$q$$ be any prime number. Then whenever $$ak$$ is divisible by $$q$$ we necessarily have that at least one of $$a, k$$ is divisible by $$q$$.

So your lemma is the case $$q = 2$$. Knowing this, we see that you could also make a copy of your proof but with $$p$$ in the role of 2 and 2 in the role of $$p$$ although it would be less intuitively appealing.

Euclid's lemma implies the uniqueness part of the Fundamental Theorem (and is mostly proved first) but of course if you have a different proof of the Fundamental Theorem, Euclid's lemma readily follows from it.

• You're right the argument is exactly the same, perhaps even simpler using the bounds on $k$: $ak=2p \implies p|k$ or $p|a$. If $p|k$ then $k=p$ and $a=2$. If $p|a$ then $k|2$ so $k=2$ and $a=p$. Both cases contradict the choice of $a$. – jskattt797 May 27 at 3:49

You can do this fairly simply with Euclid's Lemma, that if $$q$$ is a prime number and $$q\mid ab$$, then either $$q\mid a$$ or $$q\mid b$$.

If $$d$$ were a divisor of $$2p$$ other than $$1$$, $$2$$, $$p$$, or $$2p$$, then $$d$$, since it's not equal to $$1$$, must be divisible by some prime. So let $$q$$ be a prime divisor of $$d$$. But now $$q\mid d\mid2p$$ implies, by Euclid's Lemma, that either $$q\mid2$$ or $$q\mid p$$, which implies either $$q=2$$ or $$q=p$$. The latter is not possible, since if we write $$d=qk$$ and let $$q=p$$, we cannot have $$k=1$$ or $$2$$, since $$d\not=p$$ or $$2p$$, nor can we have $$k\gt2$$ since a divisor cannot be larger than the number it divides. So $$q=2$$. Thus $$d$$ can only be a power of $$2$$, and since $$d\not=2$$, it must be divisible by $$4$$. But since $$p$$ is an odd prime, we have $$p=2n+1$$ for some $$n$$, in which case $$2p=4n+2$$, which is not divisible by $$4$$. The contradiction tells us that there is no divisor of $$2p$$ other than $$1$$, $$2$$, $$p$$, and $$2p$$.

Remark: Laying out the logic of this was a little more complicated than I initially anticipated. It might be possible to compress the argument, but I don't offhand see any suitably slick way to do so. Maybe someone else does.

• $q|d|2p \implies q|2$ or $q|p$ i.e. prime $q=2$ or $q=p$. And our choice of $d$ cannot be divisible by $p$ as you noted so $q=2$, so $d=2k | 2p$ for some positive integer $k$, so $k | p$, so $k=1$ ($d=2$) or $k=p$ ($d=2p$), both of which contradict our choice of $d$. – jskattt797 May 27 at 4:08
• But doesn't "$d$, since it's not equal to $1$, must be divisible by some prime" rely on the fundamental theorem of arithmetic? – jskattt797 May 27 at 4:11
• @jskattt797, nice alternative. Regarding the "$d$ must be divisible by some prime," no, that's more basic than the fundamental theorem. Or if you like, it's the "easy" part of the fundamental theorem, which says that every integer greater than $1$ can be written as a product of primes (the "easy" part) in exactly one way (the "hard" part). – Barry Cipra May 27 at 8:02


pi notation identities

↦ If x, y, and z are the three angles of any triangle, i.e. In the language of modern trigonometry, this says: Ptolemy used this proposition to compute some angles in his table of chords. Terms with infinitely many sine factors would necessarily be equal to zero. e this identity is established it can be used to easily derive other important identities. sin {\displaystyle \theta '} The tangent (tan) of an angle is the ratio of the sine to the cosine: If the sine and cosine functions are defined by their Taylor series, then the derivatives can be found by differentiating the power series term-by-term. = ⋅ (−) ⋅ (−) ⋅ (−) ⋅ ⋯ ⋅ ⋅ ⋅. Apostol, T.M. Dividing this identity by either sin2 θ or cos2 θ yields the other two Pythagorean identities: Using these identities together with the ratio identities, it is possible to express any trigonometric function in terms of any other (up to a plus or minus sign): The versine, coversine, haversine, and exsecant were used in navigation. Algebra Calculator Calculate equations, ... \pi: e: x^{\square} 0. Finite summation. Dividing all elements of the diagram by cos α cos β provides yet another variant (shown) illustrating the angle sum formula for tangent. The always-true, never-changing trig identities are grouped by subject in the following lists: Published online: 20 May 2019. The only difference is that we use product notation to express patterns in products, that is, when the factors in a product can be represented by some pattern. When the series θ Here, we’ll present the notation with some applications. {\displaystyle \lim _{i\rightarrow \infty }\cos \theta _{i}=1} θ In trigonometry, the basic relationship between the sine and the cosine is given by the Pythagorean identity: sin2θ+cos2θ=1,{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1,} where sin2θmeans (sin θ)2and cos2θmeans (cos θ)2. The veri cation of this formula is somewhat complicated. 1 Sine, cosine, secant, and cosecant have period 2π while tangent and cotangent have period π. Identities for negative angles. None of these solutions is reducible to a real algebraic expression, as they use intermediate complex numbers under the cube roots. The ratio of these formulae gives, The Chebyshev method is a recursive algorithm for finding the nth multiple angle formula knowing the (n − 1)th and (n − 2)th values. ) These are sometimes abbreviated sin(θ) andcos(θ), respectively, where θ is the angle, but the parentheses around the angle are often omitted, e.g., sin θ andcos θ. Definition and Usage. The sum and difference identities can be used to derive the double and half angle identities as well as other identities, and we will see how in this section. cos Pi is the symbol representing the mathematical constant , which can also be input as ∖ [Pi]. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. General Identities: Summation. {\displaystyle \mathrm {SO} (2)} lim θ Purplemath. General Mathematical Identities for Analytic Functions. , [35] Suppose a1, ..., an are complex numbers, no two of which differ by an integer multiple of π. These can be "trivially" true, like "x = x" or usefully true, such as the Pythagorean Theorem's "a 2 + b 2 = c 2" for right triangles.There are loads of trigonometric identities, but the following are the ones you're most likely to see and use. You describe is supposed to end problem is not strictly a Pi Notation words that! For any measure or generalized function or basic trigonometric functions of θ also involving lengths! To a real algebraic expression, as it involves a limit and a power outside of Pi...... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance Weight Time veri cation this! Complex numbers under the cube roots months ago and quadrature components words that! Have a more concise Notation for the factorial operation numerical value they intermediate! Of an angle are sometimes referred to as the primary trigonometric functions the primary trigonometric.... Using the unit imaginary number i satisfying i2 = −1, 1.! Let i = √−1 be the imaginary unit and let ∘ denote composition of differential operators and sums than. Unit imaginary number i satisfying i2 = −1, these follow from the angle addition and theorems., commonly called trig, in pre-calculus strictly a Pi Notation words - that is, words to... 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Reducible to a real algebraic expression, we deduce Evaluate functions Simplify, as the primary functions...

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# Is $\sum\limits_{k=1}^{n-1}\binom{n}{k}x^{n-k}y^k$ always even?

Is $$f(n,x,y)=\sum^{n-1}_{k=1}{n\choose k}x^{n-k}y^k,\qquad\qquad\forall~n>0~\text{and}~x,y\in\mathbb{Z}$$ always divisible by $2$?

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Having $100$% $\LaTeX$ on the title isn't a good idea. –  Git Gud May 24 '13 at 6:35

(Hint) An odd number raised to any power is odd, and an even number raised to any power is even. In particular, $$(x + y)^n \equiv (x + y) \pmod 2$$ Using this along with the binomial formula, you should be able to prove the result.

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Thank you, Goos. –  Trancot May 24 '13 at 6:43
Yes, but your congruence is weak. It should be $(x+y)^n\equiv (x^n+y^n)\pmod{2}$, correct? –  Trancot May 24 '13 at 6:49
@BarisaBarukh that is true as well (in fact that is true for any prime). However, in the particular case of mod 2, I usually just replace all $n$th powers with the number itself, because it makes things simpler. –  Goos May 24 '13 at 6:56
But you are correct that using $(x + y)^n \equiv x^n + y^n$ makes the proof more direct. –  Goos May 24 '13 at 6:58

Hint: Recall binomial formula $$(x+y)^n=\sum\limits_{k=0}^n{n\choose k} x^{n-k} y^k$$

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Yes, I know that. This is how I came to the above result. –  Trancot May 24 '13 at 6:37
Ok do you know anything about Frobenius automorphism? –  Norbert May 24 '13 at 6:40
I've used Frobenius in an elementary differential equations course. –  Trancot May 24 '13 at 6:41
I might be able to manage if you can explain it clearly. –  Trancot May 24 '13 at 6:42
bin-o-mail? That sounds like an automatic mail deleter! :-) –  Asaf Karagila May 24 '13 at 6:43

In fact, Goos and Norbert have given the answer. (And you should also assume $n\in \mathbb{N}$)

$$f(n,x,y) = (x+y)^n - x^n -y^n$$

If

• both $x$ and $y$ are even: even - even - even = even;

• both $x$ and $y$ are odd: even - odd - odd = even;

• $x$ is even and $y$ is odd: odd - even - odd = even;

• $y$ is odd and $x$ is even: just like the case above.

So, $f(n,x,y)$ is always even.

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Try setting $y=-x+a$ and expand the powers.

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# Finding a function, given its derivative.

At the start of day zero during the summer, the temperature is $y(0) = 15$ degrees Celsius. Over a 50 day period the temperature increases according to the rule $$y'(t) = {y(t) \over 50}$$. With time $t$ measured in days. Find the formula for $y$

I not sure how to start here. Can I integrate $y'(t)$ to get back $y(t)$ ? Would it have to be a definite integral from zero to fifty ? Or and indefinite? $$\int_{0}^{50} {1\over50} dy$$ ??

• Separate the variables then integrate – randomgirl Apr 21 '16 at 13:00

Let's rewrite $y'(t)$ as $\dfrac{dy}{dt}$ and just write $y(t)$ as $y$. Then we have: $$\frac{dy}{dt} = \frac{1}{50} y$$

Separate variables to get: $$\frac{dy}{y} = \frac{1}{50} \, dt$$

Integrate both sides: \begin{align} \int \frac{dy}{y} &= \int \frac{1}{50} \, dt\\[0.3cm] \ln |y| &= \frac{1}{50} t + C\\[0.3cm] |y| &= e^{(t/50) + C}\\[0.3cm] y &= Ce^{t/50} \end{align}

Now use the fact that $y(0) = 15$ to find $C$.

Hint. Note that: $$y'-\frac{1}{50}y=0\\ e^{-t/50}y'-\frac{1}{50}e^{-t/50}y=0\\ \big(e^{-t/50}y(t)\big)'=0\\ e^{-t/50}y(t)=\text{constant}$$

• The product rule for differentiation is probably the best way to solve this for someone who's not used to the tools of differential equations (and since the asker is stumped by this question, I would assume that that is the case). Although going from the second to third line by multiplying with $e^{-t/50}$ might seem like some sort of magic trick pulled up from a hat, that's what makes it work in the end. – Arthur Apr 21 '16 at 13:03

Like so: $\frac{dy}{y}=\frac{dt}{50}$ Now integrate both sides.


+0

# Pleaze help

0
288
15
+48

The following cards are dealt to three people at random, so that everyone gets the same number of cards. What is the probability that everyone gets a red card?

[red 1] [red 2] [red 3] [yellow 1] [yellow 2] [yellow 3] [blue1] [blue 2] [blue 3]

Jan 17, 2022

#1
+2402
+2

Solution:

There are $$\large \dbinom{9}{3} = 84$$ ways to deal nine (9) cards to three (3) persons.

Give a red card to each person, then there are $$\large \dbinom{6}{3} = 20$$ ways to ways to deal the remaining six (6) cards to three (3) persons.

Probablity:

$$\Large \dfrac{\dbinom{6}{3}}{\dbinom{9}{3}} = \dfrac {5}{21} \approx 23.81\%$$

GA

--. .-

Jan 17, 2022
edited by GingerAle  Jan 17, 2022
#2
+2403
+2

Nice solution, however I also tried doing this problem and got a different answer.

I viewed all the cards as different, so the number of ways to distribute the cards was 9!/3!/3!/3! = 1680.

I think 9C3 is the number of ways to give out 3 cards to one person, not all 3.

SImilar to what you did, I then assumed everyone recieved a red card, 6 ways of doing this.

The number of ways to distribute the remaining 6 cards was 6!/2!/2!/2! = 90.

90*6/1680 = 32.14%

=^._.^=

catmg  Jan 17, 2022
#5
+2402
+1

My solution above is WRONG.

I’ve made train wrecks of these types of problems before.

I didn’t realize I’d wrecked the train this time until I read Catmg’s comment:

I think 9C3 is the number of ways to give out 3 cards to one person, not all 3.

That is true, and that means this question requires Hypergeometric distribution counts and NOT just Binomial distribution counts to correctly solve it.

The main difference between Hypergeometric distribution and Binomial distribution is that in Binomial distributions the samples sets are replaced before the next sample is drawn; in Hypergeometric distributions the samples are not replaced before the next sample is drawn.

This stands to reason: while any of the Binomial sets nCr(9,3) can exist, once a set is given to a person, the number and quality of sets remaining is greatly limited. For example: if person one receives three blue cards then person two cannot receive two yellows and a blue because there are no blue cards remaining to give.

Here is a solution using Hypergeometric distribution:

The number of possible dealt sets is  $$N=\dbinom{9}{3} * \dbinom{6}{3} * \dbinom{3}{3} = 1680 \\$$

For person one, there are  $$\dbinom{3}{1}$$ ways to select one (1) of the three (3) red cards and then there are $$\dbinom{6}{2}$$  ways to choose two more (non red) cards.

For person two, there are  $$\dbinom{2}{1}$$ ways to select one (1) of the two (2) remaining red cards and then there are $$\dbinom{4}{2}$$  ways to choose two more (non red) cards.

For person three, select the remaining red card and the two remaining (non red) cards. There is one (1) way to do this.

Then

$$n = 3 \dbinom{6}{2} \cdot 2\dbinom{4}{2} \hspace {1em} \small | \text{ Where n = the number of hands with a red card.}\\ \Large \rho_{\small \text{(3 persons with one red card)}} \normalsize = \dfrac {n} {N} = 3! \cdot \dfrac{\dbinom{6}{2}\dbinom{4}{2}}{\dbinom{9}{3}\dbinom{6}{3}} = \dfrac{9}{56} \approx 16.07\%\\$$

GA

--. .-

GingerAle  Jan 19, 2022
edited by GingerAle  Jan 19, 2022
#8
+2403
0

However, at the very end I got the answer 9/28 which is about 32.14%

6C2 = 15

4C2 = 6

9C3 = 84

6C3 = 20

3! = 6

(6*15*6)/(84*20) = (3*3*3)/(84) = 9/28

=^._.^=

catmg  Jan 20, 2022
#9
+2402
+1

Here is a much easier way to solve this. The method does not use Hypergeometric selection

There are $$(3^3) = 27$$ arrangements of success. Divide the number of successes by the total number of sets giving $$(3^3) / (nCr(9,3) = \dfrac {9}{28} \approx 32.14\%$$

I’m glad you are paying attention to my presentations and train wrecks!

If my train was loaded with toxic chemicals, I could wipe out a city. Or, in this case, at least send a student down the wrong track.

The calculation is exactly twice the value I presented for the Hypergeometric solution.

The formula is correct, but I must have made a mistake in the calculator input. I always check complex equations three times, but still, it escapes me...

I saved the input used for the calculation: (3!*nCr(6,2)*nCr(3,2))/((nCr(9,3)*nCr(6,3))

...I used a three (3) instead of a four (4) in the second binomial.

[My great uncle Cosmo was a locomotive engineer (and an electrical engineer), he would have flunked me for me for my train driving skills.]

-----------

Here’s a graphic, demonstrating the arrangements for the above solution.

$$\hspace {.3em}\left[ {\begin{array}{ccc} \scriptsize \hspace {.3em} P_1 & \scriptsize P_2 & \scriptsize P_3 \hspace {.2em} \\ \end{array} } \right] \small \hspace {.2em} \text {Persons Horizontal, sets Vertical. }\small \text{ Though identified by number, persons are indistinguishable. }\\ \left[ {\begin{array}{ccc} R & R & R \\ X & X & X \\ X & X & X \\ \end{array} } \right] \text {First arrangement of “R} \scriptsize{s} \normalsize \text {" where“X” is any other card}\\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & R \\ X & X & X \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & X \\ X & X & R \\ \end{array} } \right] \text {Third arrangement} \\ \left[ {\begin{array}{ccc} R & X & R \\ X & R & X \\ X & X & X \\ \end{array} } \right] \text {Fourth arrangement} \\ \, \\ \textbf {. . .} \hspace {1em} \textbf {. . .} \hspace {1em} \textbf {. . .} \\ \, \\ \left[ {\begin{array}{ccc} X & X & X \\ X & X & X \\ R & R & R \\ \end{array} } \right] \text {27^{th} arrangement} \\$$

From this graphic, it’s easy to see the (3^3) = 27 arrangements of success.

GA

--. .-

GingerAle  Jan 20, 2022
edited by GingerAle  Jan 20, 2022
edited by GingerAle  Jan 20, 2022
#11
+2403
0

That is a very cool way to visualize the problem.

One thing I love about math is how many different ways there are to solve the question. :))

=^._.^=

catmg  Jan 20, 2022
#3
+48
+1

GingerAle and Catmg:

You have a denominator of 10x9x8x10 because that is the total amount of combinations.

The numerator is 6 because their are 3! ways to arange the 3 white balls.

<(-_-)>_It'sFree_>(-_-)<

Jan 17, 2022
#6
+2402
+1

WTFF!! Your solution may be free, but it’s also absurd and moronic. (Just because It’s Free doesn’t mean you should be smoking it!)

The only white balls I see at the moment are in a specimen jar of formalin solution on a shelf in a cabinet. They’ve been in the specimen jar for over 21 years, and they’ve not changed much over that time.  The balls are kitty balls and were originally attached to my cat, known on the forum by his stage name, D.C. Thomas Copper.  I sometimes called him the balless wonder.  My cat was not amused by this, but my dog, Mr. Peabody, thought it quite funny. They were always busting each other’s balls. Mr. Peabody had the advantage though, because he knew where D.C. Thomas’ balls are.

D.C. Copper is now 22-years-old, still alive and somewhat active; although he mostly cat-naps. Occasionally, after his dinner, he reviews the forum and offers suggestions for potential ball-busting troll posts.   ...Like this one

GA

--. .-

GingerAle  Jan 19, 2022
edited by GingerAle  Jan 19, 2022
edited by GingerAle  Jan 19, 2022
#15
+48
+2

GA,

I am only 13 and don't smoke.

Interesting story though.  :)

I am refering to Web2.0calc being free to use!

<(it'sfree)>

ItsFree  Jan 21, 2022
#4
+117872
+2

Assume all are a bit different.

I will assume that ever person is unique.   I will also (just for the working) assume every card is unique

There are 9! ways to line up the cards but for each of those there are  3! ^3 doubling up  9!/(3!)^3 = 1680 ways total sample space

There are 3! ways to give the reds one to each person then   6!/2^3 = 90

90*3! = 540 ways for the desirable outcome

540/1680 = approx 32%

I don't know if it is right or not.   My answer seems to be in line with Catmg's.  :)  It also seems to be a reasonable answer.

I would  like to know why ItsFree is so certain their answer is correct though.  It is a bold statement to make.

Jan 18, 2022
#7
+117872
+1

It seems that we need Alan to run a Montecarlo Simulation test here.

That would give a definitive outcome.  :)

Jan 20, 2022
#10
+117872
+1

So are we all, well, me, Catmg and Ginger all in agreement that the prob is 9/28?

Melody  Jan 20, 2022
#12
+33151
+2

I've just run a Monte-Carlo simulation with 20000 trials and get a probability of 0.32125 (this will vary slightly each time it's calculated because of the use of random numbers of course).  This is close to 9/28 (≈0.32143).

Alan  Jan 21, 2022
edited by Alan  Jan 21, 2022
#13
+117872
+1

Thanks very much Alan, It is nice to get confirmation that our answers are most likely correct. :))

Melody  Jan 21, 2022
#14
+48
+2

Thank you all for helping me out.  :)

Thanks to Alan as well for doing the problem out.  ;)

<(it's_free)>

Jan 21, 2022
edited by ItsFree  Jan 21, 2022


# Two Urns Probability

Question: Suppose there is a room filled with urns of two types. Type I urns contain 5 blue balls and 5 red balls. Type II urns contain 2 red balls and 8 blue balls. There are 700 Type I urns and 300 Type II urns in the room. They are distributed randomly and look alike. An urn is selected at random from the room and a ball is drawn from it.

A) What is the probability that the urn is Type I?

So that will be: total type 1 urns/total urns $= 700/1000 = 0.7$

B) What is the probability that the ball drawn is red?

I'm confused with this part of the question. My answer is:

(type 1 red balls) $\times$ (type 2 red balls)

(5 red balls/10 total balls) $\times$ (2 red balls/10 total balls) $= 1/10$

Is $1/10$ the probability to draw a red ball correct?

• You might consider how many total balls are there and how many are red. – Cardinal Feb 24 '18 at 2:39
• Would you be able to show me how to set it up please? – Hardeep Chohan Harry Feb 24 '18 at 2:41
• I provided an answer below. – Cardinal Feb 24 '18 at 3:22
• Thanks for your answer really appreciated – Hardeep Chohan Harry Feb 24 '18 at 3:40

Type 1 Urn:

700 Total Urns (70%) 50% Chance Red

Type 2 Urn:

300 Total Urns (30%) 20% Chance Red

The Urns are indistinguishable, as are the balls. Therefore, when selecting a ball, there is a 70% chance the urn is Type 1 with a 50% chance of leading to a Red. There is a 30% chance the urn is Type 2 with a 20% chance of being a Red.

To find the total probability of drawing a red ball, you can take 70% $\times$ 50% $+$ 30% $\times$ 20%

$0.35+.06=0.41$ or 41%

700 Urns x 5 Red balls per urn = 3500 Red 300 Urns x 2 Red balls per urn = 600 Red

4100 Red 10,000 Total

$\frac{4100}{10000} = 0.41$ or 41%

Think of this as a weighted average.

The probability of a type I urn is $0.7$

The probability of then selecting a red ball is $\frac{1}{2}$

The probability of a type II urn is $0.3$

The probability of then selecting a red ball is $\frac{1}{5}$

Thus, the desired probability is $$\left(0.7\cdot\frac{1}{2}\right)+\left(0.3\cdot\frac{1}{5}\right)=0.41$$


# Expected number of objects chosen by both A and B

Suppose that A and B each randomly, and independently, choose 3 of 10 objects. Find the expected number of objects chosen by both A and B.

solution: Let $$X$$ be the number of objects chosen by both A and B. For $$1≤i≤10$$, let $$X_i=1$$ if object $$i$$ is chosen by A and B, else $$0$$ otherwise

Then $$X=X_1+\ldots+X_{10}$$.

We find $$E[X_i]=0⋅P(X_i=0)+1⋅P(X_i=1)=P(X_i=1)=9/100$$.

By the linearity of expectation, $$E[X]=10⋅E[Xi]=0.9$$

I understand this approach but I can't get the same answer by multiplying the probability of A and B having 1,2,3 objects in common by the values 1,2,3 and adding them, why is this approach concerned wrong ? I assumed x is the number of objects chosen by both thus it takes the values 1,2,3

• What do you get for the probabilities of 1, 2, or 3 objects in common? – paw88789 Feb 22 at 18:06
• The alternate approach you suggest can be correct (though it is often considered far less elegant and is more prone to error or difficulty). The error must be in your calculations themselves, which you have not yet shared. Show us your calculations and we'll tell you where your mistake is. – JMoravitz Feb 22 at 18:09
• $\frac{1}{\binom{10}{3}*\binom{10}{3}}\left [ 1*\binom{10}{1}\binom{9}{2}\binom{9}{2} + 2*\binom{10}{2}\binom{8}{1}\binom{8}{1} + 3*\binom{10}{3} \right ]$ I think I'm over counting somewhere here – Saratcı Feb 22 at 18:56
• @Saratci : disallow selection of the same "uncommon" item. $$\dfrac{\dbinom{10}{1}\dbinom{9}{2}\dbinom{7}{2}+2\dbinom{10}{2}\dbinom 81\dbinom 71+3\dbinom{10}3}{\dbinom{10}3^2}=\dfrac 9{10}$$ – Graham Kemp Feb 23 at 3:53
• Yes you are right, Thanks ! – Saratcı Feb 23 at 22:15


Jacobi's equality between complementary minors of inverse matrices

What's a quick way to prove the following fact about minors of an invertible matrix $A$ and its inverse?

Let $A[I,J]$ denote the submatrix of an $n \times n$ matrix $A$ obtained by keeping only the rows indexed by $I$ and columns indexed by $J$. Then

$$|\det A[I,J]| = | (\det A) \det A^{-1}[J^c,I^c]|,$$ where $I^c$ stands for $[n] \setminus I$, for $|I| = |J|$. It is trivial when $|I| = |J| = 1$ or $n-1$. This is apparently proved by Jacobi, but I couldn't find a proof anywhere in books or online. Horn and Johnson listed this as one of the advanced formulas in their preliminary chapter, but didn't give a proof. In general what's a reliable source to find proofs of all these little facts? I ran into this question while reading Macdonald's book on symmetric functions and Hall polynomials, in particular page 22 where he is explaining the determinantal relation between the elementary symmetric functions $e_\lambda$ and the complete symmetric functions $h_\lambda$.

I also spent 3 hours trying to crack this nut, but can only show it for diagonal matrices :(

Edit: It looks like Ferrar's book on Algebra subtitled determinant, matrices and algebraic forms, might carry a proof of this in chapter 5. Though the book seems to have a sexist bias.

• Wow - I did not know that algebra proofs could have a "sexist bias". I am too curious to let it pass --- what do you mean exactly? – Federico Poloni Feb 8 '12 at 10:19
• I was just referring to the preface, where he said the book is suitable for undergraduate students, or boys in their last years of school. Maybe the word "boy" has a gender neutral meaning back then? – John Jiang Feb 8 '12 at 19:39

The key word under which you will find this result in modern books is "Schur complement". Here is a self-contained proof. Assume $I$ and $J$ are $(1,2,\dots,k)$ for some $k$ without loss of generality (you may reorder rows/columns). Let the matrix be $$M=\begin{bmatrix}A & B\\\\ C & D\end{bmatrix},$$ where the blocks $A$ and $D$ are square. Assume for now that $A$ is invertible --- you may treat the general case with a continuity argument. Let $S=D-CA^{-1}B$ be the so-called Schur complement of $A$ in $M$.

You may verify the following identity ("magic wand Schur complement formula") $$\begin{bmatrix}A & B\\\\ C & D\end{bmatrix} = \begin{bmatrix}I & 0\\\\ CA^{-1} & I\end{bmatrix} \begin{bmatrix}A & 0\\\\ 0 & S\end{bmatrix} \begin{bmatrix}I & A^{-1}B\\\\ 0 & I\end{bmatrix}. \tag{1}$$ By taking determinants, $$\det M=\det A \det S. \tag{2}$$ Moreover, if you invert term-by-term the above formula you can see that the (2,2) block of $M^{-1}$ is $S^{-1}$. So your thesis is now (2).

Note that the "magic formula" (1) can be derived via block Gaussian elimination and is much less magic than it looks at first sight.

• I guess for any thesis involving minors the Schur complement formula would be among the first things to try. – John Jiang Feb 8 '12 at 10:30

This is nothing but the Schur complement formula. See my book Matrices; Theory and Applications, 2nd ed., Springer-Verlag GTM 216, page 41.

Up to some permutation of rows and columns, we may assume that $I=J=[1,p]$. Let us write blockwise $$A=\begin{pmatrix} W & X \\\\ Y & Z \end{pmatrix}.$$ Assume WLOG that $W$ is invertible. On the one hand, we have (Schur C.F) $$\det A=\det W\cdot\det(Z-YW^{-1}X).$$ Finally, we have $$A^{-1}=\begin{pmatrix} \cdot & \cdot \\\\ \cdot & (Z-YW^{-1}X)^{-1} \end{pmatrix},$$which gives the desired result.

These formulas are obtained by factorizing $A$ into $LU$ (namely, $L= \begin{pmatrix} I_* & 0 \\ YW^{-1} & I_* \end{pmatrix}$ and $U = \begin{pmatrix} W & X \\ 0 & Z-YW^{-1}X \end{pmatrix}$, with the $I_*$ being identity matrices of appropriate size).

• Your answer is equally valid. Thanks! – John Jiang Feb 8 '12 at 10:27

Not all has been said about this question that is worth saying -- at the very least, someone could have written down the version without the absolute values; but more importantly, there are various other equally good proofs.

Notations and statement

Let me first state the result with proper signs and no absolute values.

Standing assumptions. The following notations will be used throughout this post:

• Let $\mathbb{K}$ be a commutative ring. All matrices that appear in the following are matrices over $\mathbb{K}$.

• Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\}$.

• For every $n\in\mathbb{N}$, we let $\left[ n\right]$ denote the set $\left\{ 1,2,\ldots,n\right\}$.

• Fix $n\in\mathbb{N}$.

• Let $S_{n}$ denote the $n$-th symmetric group (i.e., the group of permutations of $\left[ n\right]$).

• If $A\in\mathbb{K}^{n\times n}$ is an $n\times n$-matrix, and if $I$ and $J$ are two subsets of $\left[ n\right]$, then $A_{J}^{I}$ is the $\left\vert I\right\vert \times\left\vert J\right\vert$-matrix defined as follows: Write $A$ in the form $A=\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq m}$; write the set $I$ in the form $I=\left\{ i_{1}<i_{2}<\cdots<i_{u}\right\}$; write the set $J$ in the form $J=\left\{ j_{1}<j_{2}<\cdots<j_{v}\right\}$. Then, set $A_{J}^{I}=\left( a_{i_{x},j_{y}}\right) _{1\leq x\leq u,\ 1\leq y\leq v}$. (Thus, roughly speaking, $A_{J}^{I}$ is the $\left\vert I\right\vert \times\left\vert J\right\vert$-matrix obtained from $A$ by removing all rows whose indices do not belong to $I$, and removing all columns whose indices do not belong to $J$.)

If $K$ is a subset of $\left[ n\right]$, then:

• we use $\widetilde{K}$ to denote the complement $\left[ n\right] \setminus K$ of this subset in $\left[ n\right]$.

• we use $\sum K$ to denote the sum of the elements of $K$.

Now, we claim the following:

Theorem 1 (Jacobi's complementary minor formula). Let $A\in\mathbb{K} ^{n\times n}$ be an invertible $n\times n$-matrix. Let $I$ and $J$ be two subsets of $\left[ n\right]$ such that $\left\vert I\right\vert =\left\vert J\right\vert$. Then,

$\det\left( A_{J}^{I}\right) =\left( -1\right) ^{\sum I+\sum J}\det A\cdot\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right)$.

Three references

Here are three references to proofs of Theorem 1:

Note that every source uses different notations. What I call $A_{J}^{I}$ above is called $A_{IJ}$ in the paper by Caracciolo, Sokal and Sportiello, is called $A\left[ I,J\right]$ in Lalonde's paper, and is called $\operatorname*{sub}\nolimits_{w\left( I\right) }^{w\left( J\right) }A$ in my notes. Also, the $I$ and $J$ in the paper by Caracciolo, Sokal and Sportiello correspond to the $\widetilde{I}$ and $\widetilde{J}$ in Theorem 1 above.

A fourth proof

Let me now give a fourth proof, using exterior algebra. The proof is probably not new (the method is definitely not), but I find it instructive.

This proof would become a lot shorter if I didn't care for the signs and would only prove the weaker claim that $\det\left( A_{J}^{I}\right) = \pm \det A\cdot\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right)$ for some value of $\pm$. But this weaker claim is not as useful as Theorem 1 in its full version (in particular, it would not suffice to fill the gap in Macdonald's book that has motivated this question).

The permutation $w\left( K\right)$

Let us first introduce some more notations:

If $K$ is a subset of $\left[ n\right]$, and if $k = \left|K\right|$, then we let $w\left( K\right)$ be the (unique) permutation $\sigma\in S_{n}$ whose first $k$ values $\sigma\left( 1\right) ,\sigma\left( 2\right) ,\ldots,\sigma\left( k\right)$ are the elements of $K$ in increasing order, and whose next $n-k$ values $\sigma\left( k+1\right) ,\sigma\left( k+2\right) ,\ldots ,\sigma\left( n\right)$ are the elements of $\widetilde{K}$ in increasing order.

The first important property of $w\left( K\right)$ is the following fact:

Lemma 2. Let $K$ be a subset of $\left[ n\right]$. Then, $\left( -1\right) ^{w\left( K\right) }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$.

You don't need to prove Lemma 2 if you only care about the weaker version of Theorem 1 with the $\pm$ sign.

Proof of Lemma 2. Let $k=\left\vert K\right\vert$. Let $a_{1},a_{2} ,\ldots,a_{k}$ be the $k$ elements of $K$ in increasing order (with no repetitions). Let $b_{1},b_{2},\ldots,b_{n-k}$ be the $n-k$ elements of $\widetilde{K}$ in increasing order (with no repetitions). Let $\gamma =w\left( K\right)$. Then, the definition of $w\left( K\right)$ shows that the first $k$ values $\gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( k\right)$ of $\gamma$ are the elements of $K$ in increasing order (that is, $a_{1},a_{2},\ldots,a_{k}$), and the next $n-k$ values $\gamma\left( k+1\right) ,\gamma\left( k+2\right) ,\ldots ,\gamma\left( n\right)$ of $\gamma$ are the elements of $\widetilde{K}$ in increasing order (that is, $b_{1},b_{2},\ldots,b_{n-k}$). In other words,

$\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots ,\gamma\left( n\right) \right) =\left( a_{1},a_{2},\ldots,a_{k} ,b_{1},b_{2},\ldots,b_{n-k}\right)$.

Now, you can obtain the list $\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( n\right) \right)$ from the list $\left( 1,2,\ldots,n\right)$ by successively switching adjacent entries, as follows:

• First, move the element $a_{1}$ to the front of the list, by successively switching it with each of the $a_{1}-1$ entries smaller than it.

• Then, move the element $a_{2}$ to the second position, by successively switching it with each of the $a_{2}-2$ entries (other than $a_{1}$) smaller than it.

• Then, move the element $a_{3}$ to the third position, by successively switching it with each of the $a_{3}-3$ entries (other than $a_{1}$ and $a_{2}$) smaller than it.

• And so on, until you finally move the element $a_{k}$ to the $k$-th position.

More formally, you are iterating over all $i\in\left\{ 1,2,\ldots,k\right\}$ (in increasing order), each time moving the element $a_{i}$ to the $i$-th position in the list, by successively switching it with each of the $a_{i}-i$ entries (other than $a_{1},a_{2},\ldots,a_{i-1}$) smaller than it.

At the end, the first $k$ positions of the list are filled with $a_{1} ,a_{2},\ldots,a_{k}$ (in this order), whereas the remaining $n-k$ positions are filled with the remaining entries $b_{1},b_{2},\ldots,b_{n-k}$ (again, in this order, because the switches have never disrupted their strictly-increasing relative order). Thus, at the end, your list is precisely $\left( a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{n-k}\right) =\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( n\right) \right)$. You have used a total of

$\left( a_{1}-1\right) +\left( a_{2}-2\right) +\cdots+\left( a_{k}-k\right)$

$=\underbrace{\left( a_{1}+a_{2}+\cdots+a_{k}\right) }_{\substack{=\sum K\\\text{(by the definition of }a_{1},a_{2},\ldots,a_{k}\text{)} }}-\underbrace{\left( 1+2+\cdots+k\right) }_{\substack{=1+2+\cdots +\left\vert K\right\vert \\\text{(since }k=\left\vert K\right\vert \text{)}}}$

$=\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right)$

switches. Thus, you have obtained the list $\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( n\right) \right)$ from the list $\left( 1,2,\ldots,n\right)$ by $\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right)$ switches of adjacent entries. In other words, the permutation $\gamma$ is a composition of $\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right)$ simple transpositions (where a "simple transposition" means a transposition switching $u$ with $u+1$ for some $u$). Hence, it has sign $\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$. This proves Lemma 2.

Exterior algebras

Now, let's introduce some more notations and state some well-known properties concerning exterior algebras.

For any $\mathbb{K}$-module $V$, we let $\wedge V$ denote the exterior algebra of $V$. The multiplication in this exterior algebra will be written as juxtaposition (i.e., we will write $ab$ for the product of two elements $a$ and $b$ of $\wedge V$) or as multiplication (i.e., we will write $a\cdot b$ for this product).

If $k\in\mathbb{N}$ and if $V$ is a $\mathbb{K}$-module, then $\wedge^{k}V$ shall mean the $k$-th exterior power of $V$. If $k\in\mathbb{N}$, if $V$ and $W$ are two $\mathbb{K}$-modules, and if $f:V\rightarrow W$ is a $\mathbb{K}$-linear map, then the $\mathbb{K}$-linear map $\wedge^{k}V\rightarrow \wedge^{k}W$ canonically induced by $f$ will be denoted by $\wedge^{k}f$. It is well-known that if $V$ and $W$ are two $\mathbb{K}$-modules, if $f:V\rightarrow W$ is a $\mathbb{K}$-linear map, then

(1) $\left( \wedge^{k}f\right) \left( a\right) \cdot\left( \wedge^{\ell}f\right) \left( b\right) =\left( \wedge^{k+\ell}f\right) \left( ab\right)$

for any $k\in\mathbb{N}$, $\ell\in\mathbb{N}$, $a\in\wedge^{k}V$ and $b\in\wedge^{\ell}V$.

If $V$ is a $\mathbb{K}$-module, then

(2) $uv=\left( -1\right) ^{k\ell}vu$

for any $k\in\mathbb{N}$, $\ell\in\mathbb{N}$, $u\in\wedge^{k}V$ and $v\in\wedge^{\ell}V$.

For any $u\in\mathbb{N}$, we consider $\mathbb{K}^{u}$ as the $\mathbb{K}$-module of column vectors with $u$ entries.

For any $u\in\mathbb{N}$ and $v\in\mathbb{N}$, and any $v\times u$-matrix $B\in\mathbb{K}^{v\times u}$, we define $f_{B}$ to be the $\mathbb{K}$-linear map $\mathbb{K}^{u}\rightarrow\mathbb{K}^{v}$ sending each $x\in\mathbb{K} ^{u}$ to $Bx\in\mathbb{K}^{v}$. This $\mathbb{K}$-linear map $f_{B}$ satisfies $\det\left( f_{B}\right) =\det B$, and is often identified with the matrix $B$ (though we will not identify it with $B$ here).

Here is another known fact:

Proposition 2a. Let $f:\mathbb{K}^{n}\rightarrow\mathbb{K}^{n}$ be a $\mathbb{K}$-linear map. The map $\wedge^{n}f:\wedge^{n}\left( \mathbb{K} ^{n}\right) \rightarrow\wedge^{n}\left( \mathbb{K}^{n}\right)$ is multiplication by $\det f$. In other words, every $z\in\wedge^{n}\left( \mathbb{K}^{n}\right)$ satisfies

(3) $\left( \wedge^{n}f\right) \left( z\right) =\left( \det f\right) z$.

Let $\left( e_{1},e_{2},\ldots,e_{n}\right)$ be the standard basis of the $\mathbb{K}$-module $\mathbb{K}^{n}$. (Thus, $e_{i}$ is the column vector whose $i$-th entry is $1$ and whose all other entries are $0$.)

For every subset $K$ of $\left[ n\right]$, we define $e_{K}\in \wedge^{\left\vert K\right\vert }\left( \mathbb{K}^{n}\right)$ to be the element $e_{k_{1}}\wedge e_{k_{2}}\wedge\cdots\wedge e_{k_{\left\vert K\right\vert }}$, where $K$ is written in the form $K=\left\{ k_{1} <k_{2}<\cdots<k_{\left\vert K\right\vert }\right\}$.

For every $k\in\mathbb{N}$ and every set $S$, we let $\mathcal{P}_{k}\left( S \right)$ denote the set of all $k$-element subsets of $S$.

It is well-known that, for every $k\in\mathbb{N}$, the family $\left( e_{K}\right) _{K\in\mathcal{P}_{k}\left( \left[ n\right] \right) }$ is a basis of the $\mathbb{K}$-module $\wedge^{k}\left( \mathbb{K}^{n}\right)$. Applying this to $k=n$, we conclude that the family $\left( e_{K}\right) _{K\in\mathcal{P}_{n}\left( \left[ n\right] \right) }$ is a basis of the $\mathbb{K}$-module $\wedge^{n}\left( \mathbb{K}^{n}\right)$. Since this family $\left( e_{K}\right) _{K\in\mathcal{P}_{n}\left( \left[ n\right] \right) }$ is the one-element family $\left( e_{\left[ n\right] }\right)$ (because the only $K\in\mathcal{P}_{n}\left( \left[ n\right] \right)$ is the set $\left[ n\right]$), this rewrites as follows: The one-element family $\left( e_{\left[ n\right] }\right)$ is a basis of the $\mathbb{K}$-module $\wedge^{n}\left( \mathbb{K}^{n}\right)$.

If $B$ is an $n\times n$-matrix and $k\in\mathbb{N}$, then evaluating the map $\wedge^{k}f_{B}$ on the elements of the basis $\left( e_{K}\right) _{K\in\mathcal{P}_{k}\left( \left[ n\right] \right) }$ of $\wedge ^{k}\left( \mathbb{K}^{n}\right)$, and expanding the results again in this basis gives rise to coefficients which are the $k\times k$-minors of $B$. More precisely:

Proposition 3. Let $B\in\mathbb{K}^{n\times n}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ n\right] \right)$. Then,

$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) = \sum\limits_{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J} ^{I}\right) e_{I}$.

(This generalizes: If $u\in\mathbb{N}$, $v \in \mathbb{N}$, $B\in\mathbb{K}^{u\times v}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ v\right] \right)$, then $\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) = \sum\limits_{I\in\mathcal{P}_{k}\left( \left[ u\right] \right) }\det\left( B_{J} ^{I}\right) e_{I}$, where the elements $e_{J}\in\wedge^{k}\left( \mathbb{K}^{v}\right)$ and $e_{I}\in\wedge^{k}\left( \mathbb{K}^{u}\right)$ are defined as before but with $v$ and $u$ instead of $n$.)

Extracting minors from the exterior algebra

Now, we shall need a simple lemma:

Lemma 4. Let $K$ be a subset of $\left[ n\right]$. Then,

$e_{K}e_{\widetilde{K}}=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }e_{\left[ n\right] }$.

Proof of Lemma 4. Let $k = \left|K\right|$. Let $\sigma$ be the permutation $w\left( K\right) \in S_{n}$ defined above. Its first $k$ values $\sigma\left( 1\right) ,\sigma\left( 2\right) ,\ldots,\sigma\left( k\right)$ are the elements of $K$ in increasing order; thus, $e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left( k\right) }=e_{K}$. Its next $n-k$ values $\sigma\left( k+1\right) ,\sigma\left( k+2\right) ,\ldots,\sigma\left( n\right)$ are the elements of $\widetilde{K}$ in increasing order; thus, $e_{\sigma\left( k+1\right) }\wedge e_{\sigma\left( k+2\right) }\wedge\cdots\wedge e_{\sigma\left( n\right) }=e_{\widetilde{K}}$.

From $\sigma=w\left( K\right)$, we obtain $\left( -1\right) ^{\sigma }=\left( -1\right) ^{w\left( K\right) }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$ (by Lemma 2).

Now, it is well-known that

$e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge \cdots\wedge e_{\sigma\left( n\right) }=\left( -1\right) ^{\sigma }\underbrace{e_{1}\wedge e_{2}\wedge\cdots\wedge e_{n}}_{=e_{\left[ n\right] }}=\left( -1\right) ^{\sigma}e_{\left[ n\right] }$.

Hence,

$\left( -1\right) ^{\sigma}e_{\left[ n\right] }=e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left( n\right) }$

$=\underbrace{\left( e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left( k\right) }\right) }_{=e_{K} }\underbrace{\left( e_{\sigma\left( k+1\right) }\wedge e_{\sigma\left( k+2\right) }\wedge\cdots\wedge e_{\sigma\left( n\right) }\right) }_{=e_{\widetilde{K}}}=e_{K}e_{\widetilde{K}}$.

Since $\left( -1\right) ^{\sigma}=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$, this rewrites as $\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }e_{\left[ n\right] }= e_K e_{\widetilde{K}}$. This proves Lemma 4.

We can combine Proposition 3 and Lemma 4 to obtain the following fact:

Corollary 5. Let $B\in\mathbb{K}^{n\times n}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ n\right] \right)$. Then, every $K\in\mathcal{P}_{k}\left( \left[ n\right] \right)$ satisfies

$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) e_{\widetilde{K} }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+k\right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] }$.

Proof of Corollary 5. Let $K \in \mathcal{P}_{k}\left( \left[ n\right] \right)$. Let $I\in\mathcal{P}_{k}\left( \left[ n\right] \right)$ be such that $I\neq K$. Then, $I\not \subseteq K$ (since the sets $I$ and $K$ have the same size $k$). Hence, there exists some $z\in I$ such that $z\notin K$. Consider this $z$. We have $z\in I$ and $z\in\widetilde{K}$ (since $z\notin K$). Hence, both $e_{I}$ and $e_{\widetilde{K}}$ are "wedge products" containing the factor $e_{z}$; therefore, the product $e_{I} e_{\widetilde{K}}$ is a "wedge product" containing this factor twice. Thus, $e_{I}e_{\widetilde{K}}=0$.

Now, forget that we fixed $I$. We thus have proven that

$e_{I}e_{\widetilde{K}}=0$ for every $I\in\mathcal{P}_{k}\left( \left[ n\right] \right)$ satisfying $I\neq K$.

Hence,

(4) $\sum\limits_{\substack{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) ;\\I\neq K}}\det\left( B_{J}^{I}\right) \underbrace{e_{I}e_{\widetilde{K}} }_{=0}=0$.

Proposition 3 yields

$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) =\sum\limits_{I\in \mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J} ^{I}\right) e_{I}$.

Multiplying both sides of this equality by $e_{\widetilde{K}}$ from the right, we find

$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) e_{\widetilde{K}} =\sum\limits_{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J}^{I}\right) e_{I}e_{\widetilde{K}}$

$=\det\left( B_{J}^{K}\right) e_{K}e_{\widetilde{K}}+\sum\limits_{\substack{I\in \mathcal{P}_{k}\left( \left[ n\right] \right) ;\\I\neq K}}\det\left( B_{J}^{I}\right) e_{I}e_{\widetilde{K}}$

$=\det\left( B_{J}^{K}\right) \underbrace{e_{K}e_{\widetilde{K}} }_{\substack{=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }e_{\left[ n\right] }\\\text{(by Lemma 4)}}}$ (by (4))

$=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] }$

$=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+k\right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] }$

(since $\left\vert K\right\vert =k$). This proves Corollary 5.

Corollary 5 is rather helpful when it comes to extracting a specific minor of a matrix $B$ from the maps $\wedge^{k}f_{B}$.

Proof of Theorem 1

Proof of Theorem 1. Set $k=\left\vert I\right\vert =\left\vert J\right\vert$. Notice that $\left\vert \widetilde{I}\right\vert =n-k$ (since $\left\vert I\right\vert =k$) and $\left\vert \widetilde{J}\right\vert =n-k$ (similarly).

Define $y\in\wedge ^{n-k}\left( \mathbb{K}^{n}\right)$ by $y=\left( \wedge^{n-k}f_{A^{-1} }\right) \left( e_{\widetilde{I}}\right)$.

The maps $f_{A}$ and $f_{A^{-1}}$ are mutually inverse (since the map $\mathbb{K}^{n\times n}\rightarrow\operatorname*{End}\left( \mathbb{K} ^{n}\right) ,\ B\mapsto f_{B}$ is a ring homomorphism). Hence, the maps $\wedge^{n-k}f_{A}$ and $\wedge^{n-k}f_{A^{-1}}$ are mutually inverse (since $\wedge^{n-k}$ is a functor). Thus, $\left( \wedge^{n-k}f_{A}\right) \circ\left( \wedge^{n-k}f_{A^{-1}}\right) =\operatorname*{id}$. Now, $y=\left( \wedge^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right)$, so that

$\left( \wedge^{n-k}f_{A}\right) \left( y\right) =\left( \wedge ^{n-k}f_{A}\right) \left( \left( \wedge^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right) \right) =\underbrace{\left( \left( \wedge ^{n-k}f_{A}\right) \circ\left( \wedge^{n-k}f_{A^{-1}}\right) \right) }_{=\operatorname*{id}}\left( e_{\widetilde{I}}\right) =e_{\widetilde{I}}$.

But (1) (applied to $V=\mathbb{K}^{n}$, $W=\mathbb{K}^{n}$, $f=f_{A}$, $\ell=n-k$, $a=e_{J}$ and $b=y$) yields

$\left( \wedge^{k}f_{A}\right) \left( e_{J}\right) \cdot\left( \wedge^{n-k}f_{A}\right) \left( y\right) =\left( \wedge^{n}f_{A}\right) \left( e_{J}y\right)$.

Thus,

$\left( \wedge^{n}f_{A}\right) \left( e_{J}y\right) =\left( \wedge ^{k}f_{A}\right) \left( e_{J}\right) \cdot\underbrace{\left( \wedge ^{n-k}f_{A}\right) \left( y\right) }_{=e_{\widetilde{I}}}$

$=\left( \wedge^{k}f_{A}\right) \left( e_{J}\right) e_{\widetilde{I}} = \left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right) e_{\left[ n\right] }$

(by Corollary 5, applied to $B=A$ and $K=I$).

Hence,

$\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right) e_{\left[ n\right] }$

$=\left( \wedge^{n}f_{A}\right) \left( e_{J}y\right) =\underbrace{\left( \det f_{A}\right) }_{=\det A}e_{J}y$

(by (3), applied to $f=f_{A}$ and $z=e_{J}y$)

(5) $=\left( \det A\right) e_{J}y$.

But (2) (applied to $\ell=n-k$, $u=e_{J}$ and $v=y$) yields

$e_{J} y = \left(-1\right)^{k \left(n-k\right)} \underbrace{y}_{=\left( \wedge^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right) } \underbrace{e_{J}} _{=e_{\widetilde{\widetilde{J}}}}$

$=\left( -1\right) ^{k\left( n-k\right) }\underbrace{\left( \wedge ^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right) e_{\widetilde{\widetilde{J}}}}_{\substack{=\left( -1\right) ^{\sum \widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }\\\text{(by Corollary 5, applied to }A^{-1}\text{, }n-k\text{, }\widetilde{I}\text{ and }\widetilde{J}\\\text{instead of }B\text{, }k\text{, }J\text{ and }K\text{)}}}$

$=\left( -1\right) ^{k\left( n-k\right) }\left( -1\right) ^{\sum \widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }$

(6) $=\left( -1\right) ^{k\left( n-k\right) +\sum\widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }$.

But

$k\left( n-k\right) +\underbrace{\sum\widetilde{J}}_{=\sum\left\{ 1,2,\ldots,n\right\} -\sum J}-\underbrace{\left( 1+2+\cdots+\left( n-k\right) \right) }_{=\sum\left\{ 1,2,\ldots,n-k\right\} }$

$=k\left( n-k\right) +\sum\left\{ 1,2,\ldots,n\right\} -\sum J-\sum\left\{ 1,2,\ldots,n-k\right\}$

$=\underbrace{\sum\left\{ 1,2,\ldots,n\right\} -\sum\left\{ 1,2,\ldots ,n-k\right\} }_{\substack{=\sum\left\{ n-k+1,n-k+2,\ldots,n\right\} \\=k\left( n-k\right) +\sum\left\{ 1,2,\ldots,k\right\} \\=k\left( n-k\right) +\left( 1+2+\cdots+k\right) }}+k\left( n-k\right) -\sum J$

$=2k\left( n-k\right) +\left( 1+2+\cdots+k\right) -\sum J$

$\equiv-\left( 1+2+\cdots+k\right) -\sum J\operatorname{mod}2$.

Hence,

$\left( -1\right) ^{k\left( n-k\right) +\sum\widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }=\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}$.

Thus, (6) rewrites as

$e_{J}y=\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }$.

Hence, (5) rewrites as

$\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right) e_{\left[ n\right] }$

$=\left( \det A\right) \left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right) e_{\left[ n\right] }$.

We can "cancel" $e_{\left[ n\right] }$ from this equality (because if $\lambda$ and $\mu$ are two elements of $\mathbb{K}$ satisfying $\lambda e_{\left[ n\right] }=\mu e_{\left[ n\right] }$, then $\lambda=\mu$), and thus obtain

$\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right)$

$=\left( \det A\right) \left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right)$.

Dividing this equality by $\left( -1\right) ^{\sum I-\left( 1+2+\cdots +k\right) }$, we obtain

$\det\left( A_{J}^{I}\right)$

$=\left(\det A\right) \dfrac{\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}}{\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right)$

$=\left( -1\right) ^{\sum I+\sum J}\det A\cdot\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right)$.

This proves Theorem 1.

I have to admit this proof looked far shorter on the scratch paper on which it was conceived than it has ended up here...

• If we agree to see a $\mathbb{K}$-matrix as a function $S\times S\to\mathbb{K}$ , with any finite index set $S$ (not just some $[m]$), then, according to Proposition 3, the matrix of the linear map $\Lambda^k(f_B)$ in the basis $\{e_I\}_{I\in\mathcal{P}_k([n])}$ is $[\det(B^I_J)]_{(I\times J)\in\mathcal{P}_k([n])\times \mathcal{P}_k([n])}$. – Pietro Majer Dec 19 '17 at 18:48
• So the functoriality of $\Lambda^k$ reads as the (generalized) Cauchy-Binet formula: en.wikipedia.org/wiki/Cauchy–Binet_formula#Generalization . In the same spirit, I'd like to see Theorem 1 as a statement on the inverse matrix of the linear map $\Lambda^k(f_A)$, or better, since it also refers to $\tilde I$ and $\tilde J$, of the linear map $\Lambda^k(f_A)^*\sim\Lambda^{n-k}(f_A)$. Can this produce a proof to Theorem 1 consisting in a plain translation of facts about exterior algebra in the language of matrices? – Pietro Majer Dec 19 '17 at 18:48
• @PietroMajer: This is what I was trying to achieve when I started writing the proof. Unfortunately, it's not directly clear to me how this works. – darij grinberg Dec 19 '17 at 19:54

Here is a short proof inspired by darij grinberg's answer and not using Schur complements (in particular, no invertibility of sub-matrices or continuity assumptions are needed).

WLOG let $I=J=[k]$. Let $A_i$ denote the $i$th column of a matrix $A$. Consider the product $$A \left[ \begin{array}{c|c|c|c|c|c} e_1 & \cdots & e_k & A^{-1}_{k+1} & \cdots & A^{-1}_n \end{array} \right] = \left[ \begin{array}{c|c|c|c|c|c} A_1 & \cdots & A_k & e_{k+1} & \cdots & e_n \end{array} \right]$$ which can also be written $$A \begin{pmatrix} I_k & A^{-1}[J,I^c] \\ 0 & A^{-1}[J^c,I^c] \end{pmatrix} = \begin{pmatrix} A[I,J] & 0 \\ A[I^c,J] & I_{n-k} \end{pmatrix}$$ Taking the determinant of both sides yields $$(\det A) (\det A^{-1}[J^c,I^c]) = \det A[I,J]$$.

• Really nice... although a nontrivial part of my answer is justifying the "WLOG" in the first sentence. Still, this simplifies a lot. – darij grinberg Feb 17 '17 at 1:19
• I believe the justification is not too bad. Assume the result is true for I=K, J=K, K=[k]. Then for any other I,J of cardinality k, we have $\det(A[I,J])=\det((PAQ)[K,K])=\det(PAQ)\det(Q^{-1}A^{-1}P^{-1}[K^c,K^c]) = \det(PAQ)\det(A^{-1}[J^c,I^c])$, where $P$ is the permutation matrix taking $I$ to $1,\dots,k$ and $I^c$ to $k+1,\dots,n$ (I believe you call this permutation $w(I)$), and $Q$ is defined similarly. So we incur a factor of $\text{sign}(P)\text{sign}(Q)$ which is $(-1)^{\sum I + \sum J}$ by a counting argument, as you proved. – user1980685 Feb 17 '17 at 15:05
• I suspect that your $Q$ is not "defined similarly" but rather is the inverse of the matrix defined similarly. But yes, this is essentially how it's done. – darij grinberg Feb 18 '17 at 5:42


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Given that Geometric series are one of the simplest examples of i… Bruno has 3 pizza stores and wants to dramatically expand his franchise nationwide. exponential functions. In finer terms, the sequence in which we multiply or divide a fixed, non-zero number, each time infinitely, then the progression is said to be geometric. Examples, solutions, videos, and lessons to help High School students learn to choose Geometric sequence sequence definition. $${S_n} = \frac{{3\left( {{2^6} – 1} \right)}}{{2 – 1}} = 3\left( {64 – 1} \right) = 189$$cm. Now that we have learnt how to how geometric sequences and series, we can apply them to real life scenarios. Example: Definition of Geometric Sequence In mathematics, the geometric sequence is a collection of numbers in which each term of the progression is a constant multiple of the previous term. change if the interest is given quarterly? How much money do View TUTORIAL 10 (APPLICATION - CHAPTER 4).pdf from MATH 015 at Open University Malaysia. $${a_1} = 3$$, $${a_n} = 96$$, $$n = 6$$, we have maximum or minimum value of the function it defines. The rabbit grows at 7% per week. monthly? Geometric Sequence & Series, Sigma Notation & Application of Geometric Series Introduction: . The geometric sequence definition is that a collection of numbers, in which all but the first one, are obtained by multiplying the previous one by a fixed, non-zero number called the common ratio.If you are struggling to understand what a geometric sequences is, don't fret! For example: If I can invest at 5% and I want $50,000 in 10 years, how much should I invest now? Try the free Mathway calculator and What will the value of an automobile be after three years if it is purchased for $$4500$$ dollars? A geometric progression (GP), also called a geometric sequence, is a sequence of numbers which differ from each other by a common ratio. Therefore, the whole length of the line is equal to $$189$$cm. This paper will cover the study of applications of geometric series in financial mathematics. What will the house be worth in 10 years? At this rate, how many boxes will When r=0, we get the sequence {a,0,0,...} which is not geometric A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio (r). If the shortest length is $$3$$cm and the longest is $$96$$cm, find the length of the whole line. Khan Academy is a 501(c)(3) nonprofit organization. This video provides an application problem that can be modeled by the sum of an a geometric sequence dealing with total income with pay doubling everyday. be rewritten as (1.151/12)12tÂ â A geometric series is the sum of the numbers in a geometric progression. You land a job as a police officer. A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. You leave the money in for 3 Try the given examples, or type in your own Please submit your feedback or enquiries via our Feedback page. Arithmetic-Geometric Progression An arithmetic-geometric progression (AGP) is a progression in which each term can be represented as the product of the terms of an arithmetic progressions (AP) and a geometric progressions (GP). TUTORIAL 10 (CHAPTER 4-APPLICATION OF ARITHMETIC AND GEOMETRIC SERIES) Linear Sequences you have in the bank after 3 years? is to sell double the number of boxes as the previous day. Educator since 2009. (GP), whereas the constant value is called the common ratio. With this free application you can: - Figure out the Nth term of the Geometric Progression given the … Each term of a geometric series, therefore, involves a higher power than the previous term. Let $${a_1} = 4500$$ = purchased value of the automobile. I have 50 rabbits. Geometric Progression is a type of sequence where each successive term is the result of multiplying a constant number to its preceding term. Copyright © 2005, 2020 - OnlineMathLearning.com. We can find the common ratio of a GP by finding the ratio between any two adjacent terms. Geometric growth is found in many real life scenarios such as population growth and the growth of an investment. problem solver below to practice various math topics. Application of geometric progression Example – 1 : If an amount ₹ 1000 deposited in the bank with annual interest rate 10% interest compounded annually, then find total amount at the end of first, second, third, forth and first years. A. Given first term (a), common ratio (r) and a integer n of the Geometric Progression series, the task is to print th n terms of the series. If the ball is dropped from 80 cm, In this tutorial we discuss the related problems of application of geometric sequence and geometric series. If the number of stores he owns doubles in number each month, what month will he launch 6,144 stores? Example: How many will I have in 15 weeks. They have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance. An “interest” problem – application of Geometric Series: Question A man borrows a loan of$1,000,000 for a house from a bank and likes to pay back in 10 years (120 monthly instalments), the first instalment being paid at the end of first month and compound interest being calculated at 6% per annum. Bouncing ball application of a geometric sequence Related Pages $$= {a_1}\left( {0.85} \right) – {a_1}\left( {0.85} \right)\left( {\frac{{15}}{{85}}} \right) = {a_1}\left( {0.85} \right)\left[ {1 – \frac{{15}}{{100}}} \right]$$ $${S_n} = \frac{{{a_1}\left( {{r^n} – 1} \right)}}{{r – 1}}$$ (As $$r > 1$$) 7% increase every year. 16,847 answers. Example 7: Solving Application Problems with Geometric Sequences. years, each year getting 5% interest per annum. $$= {a_1}\left( {0.85} \right)\left( {1 – 0.85} \right) = {a_1}\left( {0.85} \right)\left( {0.85} \right) = {a_1}{\left( {0.85} \right)^2}$$, Similarly, the value at the end of the third year The geometric series is a marvel of mathematics which rules much of the natural world. GEOMETRIC SERIES AND THREE APPLICATIONS 33 But the sum 9 10 + 9 100 + 9 1000 + 9 is a geometric series with rst term a = 10 and ratio r = 1 10.The ratio r is between 1 and 1, so we can use the formula for a geometric series: The 3rd and the 8th term of a G. P. are 4 and 128 respectively. There are many uses of geometric sequences in everyday life, but one of the most common is in calculating interest earned. fare of taxi, finding multiples of numbers within a range. In the following series, the numerators are … they sell on day 7? Estimate the student population in 2020. Geometric series have applications in math and science and are one of the simplest examples of infinite series with finite sums. Compounding Interest and other Geometric Sequence Word Problems. Your salary for the first year is $43,125. Their daily goal Use properties of exponents (such as power of a power, Geometric series are used throughout mathematics. Speed of an aircraft, finding the sum of n terms of natural numbers. Ashley Kannan. The geometric series is a marvel of mathematics which rules much of the world. Remember these examples Application of a Geometric Sequence. a. An arithmetic sequence has a common difference of 9 and a(41) = 25. Solution: This chapter is for those who want to see applications of arithmetic and geometric progressions to real life. Write the equation that represents the house’s value over time. Example: I decide to run a rabbit farm. explain properties of the quantity represented by the expression. etc.) A geometric sequence is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by r. The common ratio (r) is obtained … Factor a quadratic expression to reveal the zeros of Solve Word Problems using Geometric Sequences. The value of an automobile depreciates at the rate of $$15\%$$ per year. How much will we end up with? For example, the sequence 2, 4, 8, 16, … 2, 4, 8, 16, \dots 2, 4, 8, 1 6, … is a geometric sequence with common ratio 2 2 2. Write a formula for the student population. Educators go through a rigorous application process, and every answer they submit is reviewed by our in-house editorial team. height from which it was dropped. For example, the expression 1.15tÂ can Growth. Geometric growth occurs when the common ratio is greater than 1, that is . Geometric series are used throughout mathematics, and they have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance. If the shortest leng In this tutorial we discuss the related problems of application of geometric sequence and geometric series. In 2013, the number of students in a small school is 284. Application of Geometric Sequence and Series. A line is divided into six parts forming a geometric sequence. Each year, it increases 2% of its value. How does this On January 1, Abby’s troop sold three boxes of Girl Scout cookies online. The sequence allows a borrower to know the amount his bank expects him to pay back using simple interest. $$= {a_1}{\left( {0.85} \right)^3}$$, Hence, the value of the automobile at the end of $$3$$ years after being purchased for $$4500$$ $$96 = 3 \times {r^5}$$ $$\Rightarrow r = 2$$, For the length of the whole line, we have 5. Solution: It is finance; however, the geometric series finds perhaps its greatest predictive We welcome your feedback, comments and questions about this site or page. Example: Bouncing ball application of a geometric sequence When a ball is dropped onto a flat floor, it bounces to 65% of the height from which it was dropped. reveal and explain specific information about its approximate Find a rule for this arithmetic … Learn more about the formula of nth term, sum of GP with examples at BYJU’S. Geometric Series A pure geometric series or geometric progression is one where the ratio, r, between successive terms is a constant. Required fields are marked *. This is another example of a geometric sequence. The value of automobile at the end of the first year When a ball is dropped onto a flat floor, it bounces to 65% of the and produce an equivalent form of an expression to reveal and B. How much will your salary be at the start of year six? Find the G. P. A. Embedded content, if any, are copyrights of their respective owners. 2,3, 4,5. You will receive In order to work with these application problems you need to make sure you have a basic understanding of arithmetic sequences, arithmetic series, geometric sequences, and geometric series. $$= 4500{\left( {0.85} \right)^3} = 4500\left( {0.6141} \right) = 2763.45$$, Your email address will not be published. b. In Generalwe write a Geometric Sequence like this: {a, ar, ar2, ar3, ... } where: 1. ais the first term, and 2. r is the factor between the terms (called the "common ratio") But be careful, rshould not be 0: 1. This video gives examples of population growth and compound interest. increase or decrease in the costs of goods. Suppose you invest$1,000 in the bank. $$= {a_1} – {a_1}\left( {\frac{{15}}{{100}}} \right) = {a_1}\left( {1 – 0.05} \right) = {a_1}\left( {0.85} \right)$$, The value of the automobile at the end of the second year B. are variations on geometric sequence. Show Video Lesson In a Geometric Sequence each term is found by multiplying the previous term by a constant. sales and production are some more uses of Arithmetic Sequence. Line is divided into six parts forming a geometric series is a sequence whereas. Before going to learn how to find the height of the world are ruled by.! Fare of taxi, finding multiples of numbers within a range in financial mathematics dramatically expand franchise. Science and are one of the automobile copyrights of their respective owners equal to  %... 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