# Why is equivalent resistance in parallel circuit always less than each individual resistor?

There are $$n$$ resistors connected in a parallel combination given below.

$$\frac{1}{R_{ev}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}.......\frac{1}{R_{n}}$$

Foundation Science - Physics (class 10) by H.C. Verma states (Pg. 68)

For two resistances $$R_{1}$$ and $$R_{2}$$ connected in parallel,

$$\frac{1}{R_{ev}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{R_{1}+R_{2}}{R_{1}R_{2}}$$ $$R_{ev}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$$

We see that the equivalent resistance in a parallel combination is less than each of the resistances.

I observe this every time I do an experiment on parallel resistors or solve a parallel combination problem. How can we prove $$R_{ev} or that $$R_{ev}$$ is less than the Resistor $$R_{min}$$, which has the least resistance of all the individual resistors?

• I have to confess that when I saw that you had exhibited the harmonic sum form ($1/R_e = 1/R_1 + 1/R_2 + \dots$), I found the question confusing because I didn't understand what you didn't get. It's worth your time to think about this until it becomes obvious, not because this problem is so important but because it will start teaching you the habit of learning things from the functional form of the relationships that appear in your studies. – dmckee --- ex-moderator kitten Sep 27 '14 at 17:15
• @dmckee I didn't quite get what you said. It'll probably help me if you'll elaborate. – user49111 Sep 27 '14 at 17:26
• With experience, you will find this fact obvious, and it is worth taking the time to ponder it until it becomes so. – dmckee --- ex-moderator kitten Sep 27 '14 at 17:34
• I rolled back this question to the original version because you shouldn't put answers, or responses to answers, in the question itself. @imakesmalltalk Feel free to post an answer of your own if you would like to offer another method of answering the question. – David Z Oct 27 '14 at 0:30
• Once you believe the 2 resistor answer, it extends to any finite number. For each resistor, imagine combining all the rest into a single resistor by the law you cite. Now the combination of the one resistor and the combination resistor is less than the one. Do this once for each of the resistors, and you have that the total combination is less than any one. – Ross Millikan Oct 27 '14 at 2:53

If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Since there is more current flowing in all the resistors than through just one resistor, then the equivalent resistance must be less than the individual resistors.

The individual resistances are all positive, so the sum $$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \,,$$ is larger than the inverse of any of the individual resistances, and that means that the inverse of the sum is necessarily smaller than any of the resistances.

No mucking around with the two-resistor form required.

We can prove it by induction. Let $$\frac{1}{R^{(n)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n}$$ Now, when $n=2$, we find $$\frac{1}{R^{(2)}_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_{eq}^{(2)} = \frac{R_1 R_2}{R_1+R_2} = \frac{R_1}{1+\frac{R_1}{R_2}} = \frac{R_2}{1+\frac{R_2}{R_1}}$$ Since $\frac{R_1}{R_2} > 0$, we see that $R^{(2)}_{eq} < R_1$ and $R^{(2)}_{eq} < R_2$ or equivalently $R^{(2)}_{eq} < \min(R_1, R_2)$.

Now, suppose it is true that $R^{(n)}_{eq} < \min (R_1, \cdots, R_n)$. Then, consider $$\frac{1}{R^{(n+1)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n} + \frac{1}{R_{n+1}} = \frac{1}{R^{(n)}_{eq}} + \frac{1}{R_{n+1}}$$ Using the result from $n=2$, we find $$R^{(n+1)}_{eq} < \min ( R_{n+1} , R^{(n)}_{eq} ) < \min ( R_{n+1} , \min (R_1, \cdots, R_n))$$ But $$\min ( R_{n+1} , \min (R_1, \cdots, R_n)) = \min ( R_{n+1} , R_1, \cdots, R_n)$$ Therefore $$R^{(n+1)}_{eq} < \min ( R_1, \cdots, R_n , R_{n+1} )$$ Thus, we have shown that the above relation holds for $n=2$, and further that whenever it holds for $n$, it also holds for $n+1$. Thus, by induction, it is true for all $n\geq2$.

• $\frac {R_1}{R_2} > 0$ not 1 – Omar Elawady Apr 4 '16 at 21:41
• Your induction method is much more mathematical than the other answers....but is there any way to solve this using $AM-GM$ inequality? – Soham Jun 3 '16 at 16:11

It might be easier to think in terms of conductance which is the inverse of resistance.

The more paths there are between A and B which conduct electricity, the greater is the amount of current which can flow - ie the greater is the conductance of the network. The total conductance is greater than that of any individual path, because each additional path always increases the amount of electricity which can be conducted, it never reduces it. In particular, the total conductance is always greater than the largest individual conductance.

Translating this back in terms of resistance R (which is the inverse of conductance S - ie R = 1/S), the total resistance is smaller than the smallest individual resistance.


Question

Find inverse, by elementary row operations (if possible), of the following matrices$$\begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}$$

Solution

To check if the inverse exist we find the determinant:We have:$$A=\begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}$$So, $$\left|A\right|=1\times 7-(-5\times 3)$$$$\Rightarrow \left|A\right|=7+15=22$$Since, $$|A|\ne0$$, hence the inverse exists  Now, in order to use elementary row operations, we may write $$A=IA.$$$$\therefore \begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}A$$$$\Rightarrow \begin{bmatrix} 1 & 3 \\ 0 & 22 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}A$$                           $$R_2 \rightarrow R_2+ 5\times R_1$$     $$\Rightarrow \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ \dfrac{5}{22} & \dfrac{1}{22}\end{bmatrix}A$$                      $$R_2 \rightarrow \dfrac{1}{22} \times R_2$$$$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} \dfrac{7 }{22} & -\dfrac{3}{22} \\ \dfrac{5}{22} & \dfrac{1}{22}\end{bmatrix}A$$                   $$R_1 \rightarrow R_1- 3\times R_2$$$$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\dfrac{1}{22} \begin{bmatrix} 7 & -3 \\ 5 & 1\end{bmatrix}A$$$$\Rightarrow I=BA$$, where $$B$$ is the inverse of $$A$$.$$B=\dfrac{1}{22} \begin{bmatrix} 7 & -3 \\ 5 & 1\end{bmatrix}$$Mathematics

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# How to approach this combinatorics problem?

You have $12$ different flavors of ice-cream. You want to buy $5$ balls of ice-cream, but you want at least one to be made of chocolate and also you don't want more than $2$ balls per flavor. In how many ways can you can choose the $5$ balls?

• Does order matter? – vrugtehagel May 12 '17 at 21:32
• @vrugtehagel No. – LearningMath May 12 '17 at 21:33
• Partial hint: If $2$ of the balls are chocolate, the other $3$ balls can be chosen in ${11\choose3}+11\cdot10$ ways. Do you see why? – Barry Cipra May 12 '17 at 21:39

Calculate the coefficient of $x^5$ in $$(x^1+x^2)(x^0+x^1+x^2)^{11}$$

• Can you please elaborate where you got this from? Thanks. – LearningMath May 12 '17 at 21:31
• $(x^1+x^2)$ chooses one or two chocolate balls. Each $(x^0+x^1+x^2)$ chooses zero, one, or two balls of that flavor. – vadim123 May 12 '17 at 21:32
• Can you provide a reference for this approach? Thank you. – LearningMath May 12 '17 at 21:52
• See the second half of this. – vadim123 May 12 '17 at 23:32

This is the same as no of solutions of $\sum_{i=1}^{12} x_i=5$ where $1≤x_1≤2$ and for the other $x_i$'s $0≤x_i≤2$ which is the same as coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$

• How is it the same as the coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$ ? Can you provide a reference where I can read more about this? – LearningMath May 12 '17 at 22:06

Decide if chocolate will be a double ($c=1$) or a single ($c=0$).

If $c=0$, take a chocolate ball. Decide how many "doubles" $d\in\{0,1,2\}$ you want and make one of $11 \choose d$ specific choices regarding which non-chocolate flavors you want. These $d$ colors and chocolate are now off limits, so choose $5-1-2d=4-2d$ colors from each of the remaining $12-d-1=11-d$ legal options.

If $c=1$, take two chocolate balls. Decide how many "non-chocolate doubles" $d\in\{0,1\}$ you want and make one of $11 \choose d$ choices regarding the non-chocolate flavors. These $d$ colors and chocolate are now off limits, so choose $5-2-2d=3-2d$ flavors from each of the remaining $12-d-1=11-d$ legal options.

$$\sum_{d=0,1,2} {11 \choose d}{{11-d} \choose {4-2d}} + \sum_{d=0,1} {11 \choose d}{{11-d} \choose {3-2d}}$$

$$= {11 \choose 2} + \sum_{d=0,1} {11 \choose d} \left[{{11-d} \choose {4-2d}} + {{11-d} \choose {3-2d}}\right]$$

which is $55 + [1*495 + 11*55]=1155$.

You can also answer this using careful counting Case by case basis splitting into exactly two pairs of flavors, one pair of flavors and no pairs.

$$C_1,P_1,P_1,P_2,P_2 = 1*12*11$$ $$C_1,P_1,P_1,P_2,P_3=1*12*11*10/2!$$ $$C_1,P_1,P_2,P_3,P_4=1*12*11*10*9/4!$$

How many of these ways lead to more than $2$ chocolates?

If $$P_1=C$$ which for the first case can happen $11$ ways, the second case can happen $11*10/2!$ ways and the last case can't happen if just $P_1=C$

What about if $P_2=C$ the first case can happen in $11$ ways and the second case can happen in 11*10/2! ways.

If $P_3,P_4=C$ then we don't get $3$ or more chocolates, so we are done! The final answer is $$12*11+12*11*10/2! +12*11*10*9/4!-11*10/2!-11-11-11*10/2!=1155$$

I think the most obvious/straightforward way to solve this is to use combination 'with repetition'/'stars and bars' and then subtract all inapplicable cases. (https://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition). $\space$The formula for this for selecting $r$ from $n$ is ${{n+r-1}\choose r}$. $\space$(I like to think of this as selecting $r$ 'items' from the $n+r-1$ 'walls' and 'items' and then letting the $n-1$ 'walls' 'fall into place.')

Dividing into cases where there is one chocolate ball and more than one.

One chocolate ball of ice cream:

$11$ flavors left to choose from. Is combination 'with repetition' with $n=11$ and $r=4$ and then subtract all inapplicable cases. (All cases - cases where there are $3$ of $1$ flavor and $1$ of another - cases where there are $4$ of one flavor):

${{11+4-1=14}\choose 4}-{11\choose 1}{10\choose 1}-{11\choose 1}=1001-110-11=880$

Two chocolate balls of ice cream:

$11$ flavors left to choose from. Is combination 'with repetition' with $n=11$ and $r=3$ and then subtract all inapplicable cases. (All cases - cases where there are $3$ of $1$ flavor):

${{11+3-1=13}\choose 3}-{11\choose 1}=286-11=275$

For a grand total of $1155$.


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# What is wrong with this proof of the union of two compact sets?

I need to show that the union of two compact sets is compact. Here is what I tried:

Let $A$ and $B$ be two compact sets. Then for every open cover $\{A_n\}_{n\in I_1}$ and open cover $\{B_m\}_{m\in I_2}$, there exists finite subcovers $\bigcup\limits_{n\in I_1}^p A_n$ such that $A\subseteq\bigcup\limits_{n\in I_1}^p A_n$ and $\bigcup\limits_{b\in I_2}^k B_m$ such that $B\subseteq\bigcup\limits_{b\in I_2}^k B_m$. Then, $A\cup B\subseteq (\bigcup\limits_{n\in I_1}^p A_n)\cup(\bigcup\limits_{b\in I_2}^k B_m)\subseteq(\bigcup\limits_{n\in I_1} A_n) \cup (\bigcup\limits_{b\in I_2} B_m)$ (the union of open covers constructed as an open cover for $A\cup B$). Thus, for ever open cover of $A\cup B,$ there exists a finite subcover, so $A\cup B$ is compact.

I thought that since the open cover I constructed was arbitrary, and so was the finite subcover, then every open cover of $A\cup B$ had a finite subcover. What is wrong with this?

• Well, what's wrong is that you don't show that any cover of $C = A \cup B$ has a finite subcover. You only consider those covers that are unions of previous discussed covers of A and B have finite subcovers. If $\Alpha$ is a cover of A and $\Beta$ is a cover of B. That shows us $\Alpha \cup \Beta$ has a subcover. But what about all the covers of $C$ that are not $\Alpha \cup \Beta$? As it turns out if $\Gamma$ is an open cover then $\Gamma$ must = $\{A\}\cup \{B\}$ for some pair of open covers, and such a statement may seem obvious, but can't be assumed without proof. Mar 31, 2017 at 23:30
• You probably got the gist of what is going on, one comment that seems relevant: if you could show that every open cover of $A \cup B$ is of the form $O_1 \cup O_2$ with $O_1$ an open over of $A$ and $O_2$ another for $B,$ then you are in business (your proof goes through). Mar 31, 2017 at 23:38

$$A\cup B \subseteq \left( \bigcup_{n \in I_1} A_n \right)\cup \left( \bigcup_{m \in I_2} B_m \right)$$

may not be a general open cover for $A\cup B$. So, it is better to start with:

Let $\{ C_n \}$ be an open cover for $A \cup B$. Then since both $A$ and $B$ are compact sets, we have $\dots$

The problem with your logic is as follows:

You are right, in that a finite subcover of $A$ and a finite subcover of $B$, together gives a finite subcover of $A \cup B$.

However, to show $A \cup B$ is compact, you are supposed to start with any arbitrary open cover of $A \cup B$. This was not done: instead, you started with open covers for $A$ and $B$ separately, and used their compactness to produce a finite subcover.

The missing step : How do you know that a subcover of $A \cup B$, is also a subcover of $A$ and a subcover of $B$? Or, how would you generate subcovers of $A$ and $B$, given one of $A \cup B$? In short : you did not start with an arbitrary subcover. If you do so, it would have to be as explicit as in the answer I am going to give below for your betterment.

Hence, the answer should be as follows:

Start from definition. Let $\{ U_\alpha\}$ be an open cover of $A \cup B$.

Since $A \cup B \subset \bigcup U_\alpha$, it follows that $A \subset \bigcup U_\alpha$, and similarly, $B \subset \bigcup U_\alpha$. That is, every cover of $A \cup B$ gives a cover of $A$ and a cover of $B$. This is the key step missing from your explanation.

Now, we have finite subcovers for each one, and you can take the union of these subcovers (which remains a subcover of the original cover) to get the result.

So, in short, when we say arbitrary, we have to be careful. It is best that you start out with what is given: In compactness, we are given an open cover of the set, you never gave yourself that, instead you started with open covers for $A$ and $B$ separately, which was not given to you by the definition of $A \cup B$ being compact.

It's all about being careful. And it's absolutely fine to mistakes. It's good you have this clarified now.

As an exercise, try and extend this to a finite union of compact sets. Do you think you can extend this to an infinite union of compact sets?

You assumed, without proving it, that every open cover of $A\cup B$ is a union of an open cover of $A$ and an open cover of $B$.

Suppose for example, that $A=[0,1]$ and $B$ is some other compact set, and $\mathcal B = \{ (0.4,1] \cup G : G\text{ is an open subset of } B. \}.$ This is an open cover of $B$. Let $C=\{ [0,0.6)\cup G : G \text{ is an open subset of } B. \}.$ That is another open cover of $B$. Neither of them is an open cover of $A,$ but $\mathcal B \cup \mathcal C$ is an open cover of $A\cup B.$ Each has a finite subset that covers $B.$ Neither has a finite subset that covers $A$.

You could just say that any open cover of $A\cup B$ is an open cover of $A$. So it has a finite subcover of $A$. Similarly find another finite subset of it that covers $B$. Then take the union of those two finite subsets. That actually makes the proof simpler. If you do it that way then there is no need to prove the proposition whose proof you omitted.
PS: The needless assumption whose proof you omitted can be proved trivially: If $\mathcal A$ is an open cover of $A\cup B$ then $\mathcal A \cup \mathcal A$ is a union of an open cover of $A$ and an open cover of $B$. However, this is a needless complication in the argument.


# Mathematical induction problem

1. Jun 6, 2012

1. The problem statement, all variables and given/known data
Hello! This is I want to prove using Mathematical Induction: $1+3+5+...+(2n-1)=n^2$. The problem is: I don`t understand very much about Mathematical Induction :(

2. Relevant equations

3. The attempt at a solution
Suppose n=1. Then 1=1. Now suppose $1+3+5+...+(2n)=(n+1)^2$. Then $1+3+5+...+2n=(1+3+5+...+(2n-1))+2n=n^2+2n=n(n+2)$.
Is this correct? If yes, how does this prove my hypothesis?

2. Jun 6, 2012

You're only working with odd numbers, so the next term wouldn't be $2n$..

3. Jun 6, 2012

This is what I should do to solve the problem $1+3+5+...+(2n-1+1)=1+3+5+...+2n$. Right? Because you always do $n+1$ in the Induction step.

4. Jun 6, 2012

### sacscale

Mathematical induction attempts to show that if the equation holds for n, then it also holds for n+1. It's easier to keep straight if you use a substitution such as n=k+1

5. Jun 6, 2012

@ DDarthVader: Yes, but you're not substituting $n + 1$ in for $2n - 1$ or anything, you're subbing it in for $n$...

6. Jun 6, 2012

Doing the substitution $n=k+1$ I've got this result:
$1+3+5+...+2k+1=(1+3+5+...+(2n-1))+2k+1$
Then
$n^2 +2k+1 = (k+1)^2+2k+1 = k^2+2k+2+2k+1 = k^2+4k+3$
But since $n=k+1$ we got
$(n-1)^2+4n-4+3 =n^2-2n+4n+1 =n^2+2n+1 = (n+1)^2$
Is this correct?

Last edited: Jun 6, 2012
7. Jun 6, 2012

### SammyS

Staff Emeritus
Your conjecture: $1+3+5+...+(2n-1)=n^2$

You've already done the base step, n = 1.

For the induction step:

Let k ≥ 1. Assume that your conjecture is true for n = k. From that, show (prove) that your conjecture is true for n = k+1 .

So you need to show that $1+3+5+\dots+(2k-1)+(2(k+1)-1)=(k+1)^2$ is true,

starting with $1+3+5+\dots+(2k-1)=k^2\ .$

Last edited: Jun 6, 2012
8. Jun 6, 2012

### azizlwl

As you see here, (n+1)th value is not equal to 2n
It is equal to (2(n+1)-1)

9. Jun 6, 2012

### sacscale

Substituting n=k+1 into 2n-1 gives 2(k+1)-1.

10. Jun 6, 2012

Doing what you told me I've got this result:
$1+3+5+...+(2n-1)=n^2$.
We assume n=k. Then we try to prove if the conjecture is true for n=k+1 and we obtain: $1+3+5+...+(2n-1)+(2n-1)=n^2$
But n=k+1
$k^2+(2(k+1)-1)=(k+1)^2$
$k^2+(2(k+1)-1)=k^2+2k+1$
$2(k+1)-1=2k+1$
$2(k+1)=2k+2$
$2(k+1)=2(k+1)$
So I proved that the conjecture is also true for n=k+1. Right?

11. Jun 6, 2012

### Villyer

Isn't it bad practice to manipulate both sides of an equation in a problem such as this?

12. Jun 6, 2012

### SammyS

Staff Emeritus
Yes, somewhat indirectly.

Rather you should do something like the following.

Assume $1+3+5+\dots+(2k-1)=k^2\ .$

Now consider:

$1+3+5+\dots+(2k-1)+(2(k+1)-1)$
$=k^2+(2(k+1)-1)$   ... because of or assumption.

$=k^2+2k+1$

$=(k+1)^2$  Which is what we needed to show for the inductive step.

It's not that what you did is particularly wrong, it's just that it's not real clear that you're not somehow assuming the very thing you should be proving.

Last edited: Jun 6, 2012
13. Jun 6, 2012

### Muphrid

It's generally frowned upon to work both sides at once, yeah. Usually, one would take $k^2 + 2(k+1)-1$ and manipulate it until it's clear the result is $(k+1)^2$. Working only in one direction ensures that no illegal operations or cancellations are done, even though it may be necessary to set both sides equal just to figure out how to do it.

14. Jun 6, 2012

Well, I think I got it now! I have a list of mathematical induction to do. I'll try to solve the exercises using what you guys told me. I'll probably be back soon. Thanks guys!

15. Jun 9, 2012

### amrah

Prove by mathematical induction that n^3-n is divisible by 2 for all positive integral values of n.


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# Construct a square given a point and two lines.

The problem:

One point of the square - $$(2,5)$$, and lines $$x=1$$, $$x=6$$ are given (1 point of the square is on each given line), all points of the square are in the first quadrant.
A square needs to be constructed with this information.

How would you solve this problem and thus determine if a square can even be constructed?
I'm interested more in your thought process than the solution.
How do you approach a problem like this?
Which steps do you take and why?

I've considered equating the distances between points (2,5) and (1, a) with the distance between (2,5) and (6,b).I don't think that anything can be done with this, maybe I have to find something else and make a system of equations?I've tried to find the distance between points (1, a) and (6,b) - that would be equal to sqrt(2)*n (n is the length of square's side), connect it with some distance (between the given and unknown point) but I can't extract one variable so I can insert it into the first equation.

• Welcome to Math.SE! Typically, the community here likes to see your thoughts about the problem, as this helps answerers target their responses to your experience level, while avoiding wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.) As you are interested more in thought process than solution, perhaps you can say something about your motivation: are you conducting research on problem-solving techniques? trying to be a better problem-solver? (doing homework that asks about process?) – Blue Nov 11 '18 at 12:00
• I've edited the question.I'm interested in general problem solving techniques because I typically solve them by doing everything that can be done which certainly isn't an efficient way of solving problems. – JoeDough Nov 11 '18 at 12:12
• Consider you found the square and draw the two horizontal lines from the vertexes not laying on the vertical lines you already have, the intersections of the four lines will give another figure, that can give many insights. – N74 Nov 11 '18 at 13:07

Let $$A$$ be the given point, and let $$B$$ and $$C$$ be points on the respective given lines. The key observation is

Because $$\overline{AB}$$ and $$\overline{AC}$$ are congruent and perpendicular, the horizontal distance between $$A$$ and $$B$$ must equal the vertical distance between $$A$$ and $$C$$, and vice-versa.

The image shows that $$\triangle ABP \cong \triangle ACQ$$, proving the observation immediately. The image also shows that there are clearly two matching choices for $$B$$ and $$C$$, so that there are two solution squares.

The idea also works when $$A$$ is not between the lines (note that how the $$B$$ and $$C$$ points match is reversed from above):

(Of course, $$A$$ on a line works, too, as an obvious special case where $$C$$ and $$C^\prime$$ coincide.) So, we see that it's always possible to construct the squares. $$\square$$

We leave as an exercise to the reader the task of finding the coordinates of the points for the specific problem in the original question.

As for thought process ...

• My first instinct is usually to abandon specific numbers and to generalize; this prevents important algebraic and geometric patterns from being lost in the arithmetic of numbers. So, instead of $$A=(2,5)$$, $$x=1$$, $$x=6$$, I wanted to consider $$A=(p,q)$$, $$x=r$$, $$x=s$$.

• These days, I have a bit of a knee-jerk impulse to jump into Mathematica to try a brute-force solution. I quickly assigned some coordinates ---$$A=(p,q)$$, $$B=(r,b)$$, $$C=(s,c)$$--- and entered conditions for making the squares: $$\overline{AB}\perp\overline{AC}$$ becomes $$(A-B)\cdot(A-C)= 0$$, while $$\overline{AB}\cong\overline{AC}$$ becomes $$(A-B)\cdot(A-B)=(A-C)\cdot(A-C)$$.

• I let Mathematica do some instantaneous symbol-crunching, which in this case yielded linear constraints on the $$y$$-coordinates of $$b$$ and $$c$$. That told me the problem was actually easy, so I took a step back. The "key observation" above came to me pretty quickly ... shaming me because I didn't think of it sooner. :) (But, hey ... I only lost about two minutes!)

• Then I went to the GeoGebra app to draw-up some diagrams that make the "key observation" obvious, and double-checking the point-outside-the-lines case. (I ofttimes go right into GeoGebra to experiment, but a first pass through Mathematica seemed quicker in this case.)

Hint:
Two parallel lines are: $$x=1$$ and $$x=6$$
distance between them is $$5$$ units.
So, you can construct a square having $$(1,5)$$ as one co-ordinate, and another lying on line $$x=6$$, opposite to $$(1,5)$$ ,i.e., $$(6,5)$$

Take other $$2$$ points on lines $$x=1$$ and $$x=6$$ at a distance of $$5$$ units from $$2$$ points already taken.

So, there are $$2$$ possible squares: $$(1,5),(6,5),(1,0),(6,0)$$ and $$(1,5),(6,5),(1,10),(6,10)$$

• I've made a mistake, first line isn't $x=2$ but $x=1$ – JoeDough Nov 11 '18 at 11:10
• no problem, just shift the points, and now length of side will be 5 – pooja somani Nov 11 '18 at 11:11
• @JoeDough is your point not (1,5) now? – pooja somani Nov 11 '18 at 11:16
• But where's the point (2,5)?It's the only given point of the square, other 2 are somewhere on the given lines and 4th one has to be found.I've edited the question to be more clear. – JoeDough Nov 11 '18 at 11:17
• ohk, then you will have to plot all points and proceed accordingly – pooja somani Nov 11 '18 at 11:18


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# If √(xy) = xy, what is the value of x + y?

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If root(xy) = xy what is the value of x + y?  [#permalink]

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11 Jul 2012, 14:05
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If $$\sqrt{xy} = xy$$ what is the value of x + y?

(1) x = -1/2
(2) y is not equal to zero

What i did was,

(XY)^1/2 = XY
XY =(XY)^2

so, I cancelled out XY and finally I got the below rephrased equation

XY=1.

BUT in the MGMAT explanation, I found that they are not cancelling out XY.

Below is their rephrased equation.

XY = (XY)^2

XY-(XY)^2=0

XY [1-(XY)] = 0

so, XY = 0 or XY = 1.

My question is why we are not cancelling out, and when we should use cancelling technique.
Math Expert
Joined: 02 Sep 2009
Posts: 49206
Re: If root(xy) = xy what is the value of x + y?  [#permalink]

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11 Jul 2012, 14:22
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If $$\sqrt{xy} = xy$$ what is the value of x + y?

$$\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.

(1) x = -1/2 --> either $$-\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\frac{1}{2}$$ OR $$-\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Not sufficient.

(2) y is not equal to zero. Clearly not sufficient.

(1)+(2) Since from (2) $$y\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Sufficient.

As for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.

Hope it's clear.
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Re: If root(xy) = xy what is the value of x + y?  [#permalink]

### Show Tags

11 Jul 2012, 22:57
Responding to a pm:

"If $$\sqrt{XY} = XY$$ what is the value of x + y?

(1) x = -1/2
(2) y is not equal to zero

What i did was,

(XY)^1/2 = XY
XY =(XY)^2

so, I cancelled out XY and finally I got the below rephrased equation

XY=1.

BUT in the MGMAT explanation, I found that they are not cancelling out XY.

Below is their rephrased equation.

XY = (XY)^2

XY-(XY)^2=0

XY [1-(XY)] = 0

so, XY = 0 or XY = 1.

My question is why we are not cancelling out and when we should use cancelling technique."

For the time being, forget this question. Look at another one.

Which values of x satisfy this equation: $$x^2 = x$$
Let's say we cancel out x from both sides. What do we get? x = 1.
So we get that x can take the value 1.

But is your answer complete? I look at the equation and I say, 'x can also take the value 0.' Am I wrong? No.
x = 0 also satisfies your equation. So why didn't you get it using algebra? It is because you canceled x.
Let me treat this equation differently now.

$$x^2 = x$$
$$x^2 - x = 0$$
$$x * (x - 1) = 0$$
x = 0 OR (x - 1) = 0 i.e. x = 1

Now I get both the possible values that x can take. I do not lose a solution.

When you cancel off a variable from both sides of the equation, you lose a solution so you should not do that. You can cancel off constants of course.
Rule of thumb: Do not cancel off variables. Take them common. In some cases, it may not matter even if you do cancel off but either ways, your answer will not be incorrect of you don't cancel. On the other hand, sometimes, your solution could be incomplete if you do cancel and that's a problem.
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Intern
Joined: 12 Jul 2012
Posts: 21
Re: If root(xy) = xy what is the value of x + y?  [#permalink]

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12 Jul 2012, 03:00
Here is my solution

Take square of both sides:
(root(xy) )^2 = (xy)^2
xy = (xy)^2
With Option (1): x = -1/2
-1/2y = 1/4 * y^2
y = 4 * (-1/2)
y = -2
Hence, X+Y = -1/2 - 2 = -5/2.
Math Expert
Joined: 02 Sep 2009
Posts: 49206
Re: If root(xy) = xy what is the value of x + y?  [#permalink]

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12 Jul 2012, 03:06
RamakantPareek wrote:
Here is my solution

Take square of both sides:
(root(xy) )^2 = (xy)^2
xy = (xy)^2
With Option (1): x = -1/2
-1/2y = 1/4 * y^2
y = 4 * (-1/2)
y = -2
Hence, X+Y = -1/2 - 2 = -5/2.

There is another value of $$y$$ apart -2 which satisfies $$-\frac{1}{2}*y=\frac{1}{4}*y^2$$, namely $$y=0$$.

Complete solution here: if-root-xy-xy-what-is-the-value-of-x-y-135646.html#p1103625
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15 Oct 2014, 20:57
1
Bunuel wrote:

Tough and Tricky questions: Algebra.

If √(xy) = xy, what is the value of x + y?

(1) x = -1/2
(2) y is not equal to 0

Statement 1: $$X=-1/2$$.... substitute in main equation

Scenario A: $$Y = -2/1$$ is a reciprocal
$$\sqrt{xy}=xy$$
LHS =$$\sqrt{-1/2*-2/1}= \sqrt{1}= 1$$
RHS =$$-1/2*-2/1= -1*-1 = 1$$
Therefore, LHS = RHS

Scenario B: $$Y = 0$$
LHS =$$\sqrt{-1/2*-0}= \sqrt{0}= 0$$
RHS =$$-1/2*0= 0$$
Therefore again, LHS = RHS

Hence, we have two possible solutions, therefore Statement 1 is insufficient

Statement 2 : $$Y$$ is not equal to zero
But, X could be equal to zero or be the reciprocal of Y, therefore, statement 2 would fall under the same scenarios as statement 1

Hence, we have two possible solutions, therefore Statement 2 is insufficient

Both Statement 1 & 2 together confirms that $$X & Y$$ both are not equal to zero, therefore, they have to be reciprocals and since statement gives us the value of $$X=-1/2$$, we can also the value of $$Y=-2/1$$ (Reciprocal of X) and thereafter we can find the value of $$X + Y = -1/2 + -2/1 = -5/2$$

Hence, both together are sufficient, therefore C is the correct Answer!

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Re: If √(xy) = xy, what is the value of x + y?  [#permalink]

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15 Oct 2014, 21:26
1
Bunuel wrote:

Tough and Tricky questions: Algebra.

If √(xy) = xy, what is the value of x + y?

(1) x = -1/2
(2) y is not equal to 0

I did something similar to DMMK, except for one thing:
I know $$\sqrt{n}$$ = $$n$$ only when $$n$$ = 0 or $$n$$ = 1. For all other values of $$n$$, the two would not be equal. Knowing this made it much simpler to plug in the statements. I just had to determine if from the statement(s) provided, can I rule out xy = 0 or xy=1.

Stament 1:
Knowing that x = -1/2, y could equal 0 or -2, and still make the premise true. Either value of y would make x + y a different value. Therefore, it is insufficient.

Statement 2:
Knowing that y is not equal to 0, x could equal zero or the inverse of y, and still make the premise true. Either value of x would make x + y a different value. Therefore, it is insufficient.

Now to evaluate both statements together.

The reason we rejected statement 1 by itself was because y could equal one of two possible values. Statement 2 eliminates one of those options. Therefore y must equal -2. And therefore both statements together are sufficient to answer "what is the value of x+y."
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Re: If √(xy) = xy, what is the value of x + y?  [#permalink]

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15 Oct 2014, 22:12
1
Bunuel wrote:

Tough and Tricky questions: Algebra.

If √(xy) = xy, what is the value of x + y?

(1) x = -1/2
(2) y is not equal to 0

let xy=k, thus
k^(1/2) =k;
squaring both sides we have
k=k^2
k(k-1)=0
k=0 or 1
i.e. xy=0 or xy=1

st.1

for x=-1/2
both y=0 and y=-2 satisfy the possible value of xy i.e. 0 or 1
hence not sufficient

st.2

y is not equal to zero. clearly not sufficient.
as nothing is said about x, therefore x can take any value it can be zero, fraction, integer etc.

combining st.1 and st.2

we know that y cannot be equal to zero. thus y=-2 and x=-1/2
and x+y= -2-(1/2)=-5/2

hence sufficient.
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Re: If root(xy) = xy what is the value of x + y?  [#permalink]

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04 Jul 2015, 08:43
Why can't x and y both be -1? Thanks
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05 Jul 2015, 08:19
ElCorazon wrote:
Why can't x and y both be -1? Thanks

From the stem x = y = -1 is possible, in this case xy = 1 (check my solution above). But then the first statement says that x = -1/2, thus these values are no longer possible.

Hope it's clear.
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Re: If √(xy) = xy, what is the value of x + y?  [#permalink]

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16 Nov 2015, 03:02
Bunuel wrote:

Tough and Tricky questions: Algebra.

If √(xy) = xy, what is the value of x + y?

(1) x = -1/2
(2) y is not equal to 0

Given: √(xy) = xy
(xy)^2 - xy = 0
xy = 0 or 1 - (i)

Required: x + y = ?

Statement 1: x = -1/2
From the given equation (i), we can have two different values of y.
Hence two different values of x + y
INSUFFICIENT

Statement 2: y is not equal to 0
No information about x
INSUFFICIENT

Combining Statement 1 and Statement 2: x = -1/2 and y is not equal to 0
From (i), xy cannot be 0 since both x and y are not = 0
Hence xy = 1
y = -2
x + y = $$-\frac{5}{2}$$
SUFFICIENT

Option C
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If √(xy) = xy, what is the value of x + y?  [#permalink]

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14 Apr 2017, 07:13
Bunuel wrote:

Tough and Tricky questions: Algebra.

If √(xy) = xy, what is the value of x + y?

(1) x = -1/2
(2) y is not equal to 0

The trick here is from $$\sqrt{xy}=xy$$ anyone could easily deduce that $$xy=1$$ or $$xy=0$$ and just consider one of these cases.

Here is my solution:
$$\sqrt{xy}=xy \implies \sqrt{xy}-xy=0 \implies \sqrt{xy}(\sqrt{xy}-1)=0$$
Hence we have $$xy=0$$ or $$xy=1$$.

(1) If $$x=-\frac{1}{2}$$, we need to consider two cases:

Case 1: If $$xy=0\implies y=0 \implies x+y = -\frac{1}{2}$$

Case 2: If $$xy=1 \implies y=-2 \implies x+y = -\frac{5}{2}$$

It's clear that (1) is insufficient.

(2) If $$y \neq 0$$, we still need to consider two cases:

Case 1: If $$xy=0\implies x=0 \implies x+y = y$$.

If $$y=1 \implies x+y = 1$$
If $$y=2 \implies x+y = 2$$

Hence, (2) is insufficient.

Now combine (1) and (2):
Since $$x \neq 0$$ and $$y \neq 0$$ hence $$xy \neq 0 \implies xy=1$$

Since $$x= -\frac{1}{2} \implies y = -2 \implies x+y = -\frac{5}{2}$$

Hence (1) & (2) are sufficient.
The answer is C.
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Re: If √(xy) = xy, what is the value of x + y?  [#permalink]

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10 Sep 2017, 13:41
Bunuel wrote:

Tough and Tricky questions: Algebra.

If √(xy) = xy, what is the value of x + y?

(1) x = -1/2
(2) y is not equal to 0

rewrite

xy= (xy)^2

St 1

Reciprocal property.

-1/2(-2) =1
[-1/2(-2)]^2 =1

insuff

St 2

Eliminating the possibility of 0 from Y still leaves the possibility of x being 0 or some other numbers which leads to several possibilities.

insuff

St 1 and St 2

Eliminates possibility of y=-2

C
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Re: If √(xy) = xy, what is the value of x + y?  [#permalink]

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27 Dec 2017, 00:35
Bunuel wrote:
If $$\sqrt{xy} = xy$$ what is the value of x + y?

$$\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.

(1) x = -1/2 --> either $$-\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\frac{1}{2}$$ OR $$-\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Not sufficient.

(2) y is not equal to zero. Clearly not sufficient.

(1)+(2) Since from (2) $$y\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Sufficient.

As for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.

Hope it's clear.

Is it safe to say : x^2 = x only when x = 0 or 1
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Re: If √(xy) = xy, what is the value of x + y?  [#permalink]

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27 Dec 2017, 00:39
QZ wrote:
Bunuel wrote:
If $$\sqrt{xy} = xy$$ what is the value of x + y?

$$\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.

(1) x = -1/2 --> either $$-\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\frac{1}{2}$$ OR $$-\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Not sufficient.

(2) y is not equal to zero. Clearly not sufficient.

(1)+(2) Since from (2) $$y\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Sufficient.

As for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.

Hope it's clear.

Is it safe to say : x^2 = x only when x = 0 or 1

Yes.

x^2 = x;

x^2 -x = 0;

x(x - 1) = 0;

x = 0 or x = 1.
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Re: If √(xy) = xy, what is the value of x + y?  [#permalink]

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27 Feb 2018, 10:18
Hi, why aren't we accounting for option where both x and y are equal to 1? then x+y is 2
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Re: If √(xy) = xy, what is the value of x + y?  [#permalink]

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27 Feb 2018, 10:27
1
krikre wrote:
Hi, why aren't we accounting for option where both x and y are equal to 1? then x+y is 2

Doesn't (1) say that x = -1/2? How it can be 1 then?
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Re: If √(xy) = xy, what is the value of x + y?  [#permalink]

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28 Feb 2018, 01:19
prakash111687 wrote:
If $$\sqrt{xy} = xy$$ what is the value of x + y?

(1) x = -1/2
(2) y is not equal to zero

What i did was,

(XY)^1/2 = XY
XY =(XY)^2

so, I cancelled out XY and finally I got the below rephrased equation

XY=1.
there are various cases that are possible as following

BUT in the MGMAT explanation, I found that they are not cancelling out XY.

Below is their rephrased equation.

XY = (XY)^2

XY-(XY)^2=0

XY [1-(XY)] = 0

so, XY = 0 or XY = 1.

My question is why we are not cancelling out, and when we should use cancelling technique.

IMO C
$$\sqrt{xy} = xy$$
1.x=0 ,y = anything
2.Y=0 , x=anything
3. both x and y 1 or -1 or x=1 and y=-1 or vise versa
4.x=1/z and y=z *any integer except 0
lets come to statement 1.
x=-1/2 y can be -2 or 0 insufficient.
2nd statement y not equal to 0 but don't know about x
combining both case with 0 is eliminated only one case left so C is the answer
Re: If √(xy) = xy, what is the value of x + y? &nbs [#permalink] 28 Feb 2018, 01:19
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# Why is $\sum\limits_{k=1}^{100} 0.01$ different than $\sum\limits_{k=1}^{100} \frac{1}{100}$?

Why does evalutating $$\sum\limits_{k=1}^{100} 0.01$$ not result in 1?

What I get is 1.0000000000000007.

I know that the number 0.01 in binary results in an infinite sequence of zeros and ones, which is truncated when stored in a memory as a floating point number. So information is lost and the result of the sum will no longer be exactly 1.

However, with $$\sum\limits_{k=1}^{100} \frac{1}{100}$$, I do get exactly 1.

Could someone explain why?

• For your own edification, compare the results of Total[Table[0.01, {100}]] and Total[Table[0.01, {100}], Method -> "CompensatedSummation"]. – J. M.'s torpor Oct 22 '18 at 6:21
• @J.M.iscomputer-less Interestingly Total seems to perform much better than Sum, even without "CompensatedSummation". It is somehow more intelligent. Consider ListLinePlot[{Table[Sum[0.01, {i, n}] - n/100, {n, 1, 100}], Table[Total[Table[0.01, {n}]] - n/100, {n, 1, 100}]}] – kirma Oct 22 '18 at 9:03

Welcome to the world of Mathematica numerics.

Mathematica offers three arithmetic systems for evaluating numerical expressions

• Numerics based on exact numbers. Such computations will always give an exact result, but require all the numbers involved to be integers or rational numbers. Your second expression qualifies as an exact computation and, thus, produces the exact result 1.

• Numerics based on arbitrary precision numbers. This allows you the precision a computation to specified value. To work, such computations require the numbers involved to be integers, rational numbers or arbitrary precision having a precision greater or equal to the specified precision.

• Numerics based on machine floating point numbers. This is used whenever a real number, which does not qualify as an arbitrary precision number, appears in a computation. Machine numerics give the fastest computation, but since there is no control over precision. For long computations precision will be lost, sometimes drastically (leading to zero precision – i.e., garbage out). Your first expression falls into this category, but retains good precision. However, when you display all the digits in the result the inexactness shows.

In you specific example, since the deviation from the exact result is small, you can recover the exact result with Rationalize

Example

Sum[.01, {100}] // FullForm

1.0000000000000007

Rationalized example

Sum[.01, {100}] // Rationalize // FullForm

1

Mathematica treats 0.01 as a machine precision floating point number, while it treats 1/100` as an exact rational quantity. Thus the first sum follows all the grody rules of floating point arithmetic that programmers have come to know and hate.

There are ways to make Mathematica use more digits, though.


Proof of obtaining multiple of 10 by using arithmetic operations on any three numbers

While doing a few math puzzles I noted that you could take any three numbers (integers) and use basic arithmetic operations (Addition, Subtraction, Multiplication and Parentheses only) to obtain a multiple of 10, using each number only once.

Eg.
Using 29, 73, 36: $73+36-29=80$
Using 2, 4, 7: $2*7-4=10$

Is there a proof for this theory in particular, or is there a more general theorem that applies to this? If not, is there a set of three numbers which disproves this theory?

I found a few clues while solving this:

If one of the numbers is a multiple of 5 (call it $x$) then there definitely exists a solution divisible by 10. This one is easily provable as you merely need an even number to multiply with. If any one of the remaining two numbers is even, you can use that to multiply with $x$ to get number divisible by 10. If both numbers are odd then you can add them to get an even number to multiply with $x$.

It would seem as only the units place digits have any bearing on whether it's a multiple of 10 or not. As you could take any of the examples, add or subtract any multiple of 10 from it and do the same operations to still get a multiple of ten. I feel this is also easily provable but I can't think of a way to do it which isn't long-winded.
EDIT: It can be proven as such, let $a,b$ be two integers s.t $a+b|10$
Adding two multiples of 10 ($10m,10n$) to $a$ and $b$ we get $10m+a+10n+b$
Since $a+b|10$ it can be written as $10m+10n+10k$ where $a+b=10k$
Which is divisible by ten
Now let $a,b$ be two integers s.t. $a.b|10$
Adding $10m,10n$ to $a$ and $b$ we get $(10m+a)(10n+b)$
$=100mn+10an+10bm+ab$ or $100mn+10an+10bm+10k$
Which is divisible by ten
Therefore any combination of Addition and Multiplication with the three no.s shouldn't matter.
I realise that could replace 10 with any no. in this proof and it would still work. So we really need to just prove for every single digit no.

I ran a program in python that checked for every combination of single digit no.s, but it didn't find any combination that disproves this.

I'm not sure how this question would be categorised, hence the lack lustre tags.

I'm fairly new to StackExchange. Please forgive me if I've worded this question poorly.

• That's a very nice observation that only the units digit has any bearing, and is essentially the underlying idea of "modular arithmetic." This is an interesting problem, and I'll be fairly surprised if it's true. However, there are $3$ ways to order $a, b, c$, a couple choices of symbols between $a,b$ and $b,c$ and with the ability to use parentheses... very interesting! If we have the three numbers $a, b, c$, are we allowed to order them any way we like? – pjs36 Nov 5 '17 at 6:47
• How do you get a multiple of $10$ this way using $1$, $2$, and $4$? – alex.jordan Nov 5 '17 at 7:58
• @alex.jordan $2\cdot (4+1)$ – Kyle Miller Nov 5 '17 at 7:59
• Oh, I see. I somehow thought only addition and subtraction were in play. Didn't read well. – alex.jordan Nov 5 '17 at 8:00
• @pjs36 From the two examples in the OP, it seems clear that reordering is permitted. – Erick Wong Nov 5 '17 at 8:07

Let's collect some of the useful observations mentioned and implicit in the OP:

1. Only the residue class of $a,b,c$ mod $10$ matters.
2. If any of $a,b,c$ is divisible by $5$ then there is a solution.
3. If any subset of $\{a,b,c\}$ has a solution, then so does $\{a,b,c\}$.

As a consequence of the above, we may assume WLOG that $\{a,b,c\}$ are distinct single digits from $1,2,3,4,6,7,8,9$.

With a little more effort, we can justify that $a$ can be replaced by $-a$, which will let us further restrict to the digits $1,2,3,4$. It's not too hard to be convinced of this by playing around with some examples, but let's give a proper proof by structural induction. More specifically, we will prove:

Proposition: Let $f(x_1,\ldots,x_n)$ be a fully parenthesized expression using each $x_i$ exactly once and only $+,-,*$ operators. Then at least one of $-f$ or $f$ can be written as a similar expression using $-x_1,x_2,\ldots,x_n$ exactly once.

This is trivial for $n=1$, and for $n=2$ we quickly verify each of the four possible expressions of two variables: $$-(a+b) = (-a)-b,\\-(a-b) = (-a)+b,\\(b-a) = b+(-a),\\-(a*b) = (-a)*b.$$

We can now do a structural induction: suppose $n\ge 3$. By extracting the highest-level operator of $f(x_1,\ldots, x_n)$, we may write $f$ as $h(g_1(\cdots),g_2(\cdots))$, where each of $g_1,g_2,h$ are valid expressions, and the arguments of $g_1, g_2$ partition the set $\{x_1,\ldots,x_n\}$. I've deliberately obscured the arguments because the notation becomes unwieldy, as the idea is better illustrated by a simple example:

If $f(x_1,x_2,x_3,x_4,x_5) = ((x_1 * x_5) + x_2) - (x_3 - x_4)$, then we take $$h(a,b) = a-b,\quad g_1 = (x_1 * x_5) + x_2,\quad g_2 = x_3 - x_4.$$

Note that $h,g_1,g_2$ each take $<n$ arguments so we could apply the inductive hypothesis to any of them as desired. Now, $x_1$ belongs to exactly one of $g_1$ or $g_2$. By adjusting $h$, we can assume WLOG that it belongs to $g_1$. By hypothesis, either $-g_1$ or $g_1$ can be written in terms of $-x_1$ and the remaining arguments of $g_1$. If it is $g_1$ that may be so written, then we are done since $f=h(g_1,g_2)$ can be written in terms of $-x_1,x_2,\ldots,x_n$. Else, apply the induction hypothesis again to $h$ to see that either $f=h(g_1,g_2)$ or $-f=-h(g_1,g_2)$ can be written in terms of $-g_1$ and $g_2$, and thus also in terms of $-x_1,x_2,\ldots,x_n$.

The above proposition lets us freely replace $9$ by $-9$ (which is equivalent to $1$), etc., and so we can narrow down the set $\{a,b,c\}$ to a subset of $\{1,2,3,4\}$. Lucky for us, there's only four such subsets:

$$(1 + 2 - 3) = 0,\\ (1 + 4)*2 = 10,\\ (1 + 3 -4) = 0,\\ (2+3)*4 = 20.$$

• I was going to write the same thing. I believe this one deserves the bounty. – user334639 Nov 5 '17 at 9:00
• Very nicely done, I probably will award the bounty to this answer (I want to keep the bounty alive to reward the nice question; OP had but a single upvote when I set the bounty). Structural induction isn't something I'm familiar with, so I will admit that it's my least favorite part. I notice that in each of the four ultimate cases, addition is used -- I wonder if we could somehow combine that with factoring out a negative, and avoid structural induction. Probably not. At any rate, very nice! – pjs36 Nov 6 '17 at 2:36
• Ah I see you've used $-4,-3,-2,-1,1,2,3,4$ instead of $1,2,3,4,6,7,8,9$ and then used my modulo(10) proof.First time I'm coming across structural induction so the proof is a bit hard to chew. Nevertheless, it's a clever proof. – Carlton Banks Nov 6 '17 at 10:02
• @CarltonBanks Thanks, to be fair this is really just an induction on the number of terms (it’s structural in the sense that arithmetic expressions aren’t built up by just “adding one more term”). It is also overkill when we’re dealing with only 3 terms, but you did express an interest in more general principles in the OP. Finally, the induction is only needed for a relatively minor technical point: it justifies that if we can find a $0$ using unary $-$ (which doesn’t appear to be allowed), then we can also find one without any unary $-$. – Erick Wong Nov 6 '17 at 18:24

If you have directly verified this for all triples of one-digit numbers, then you have proved it. Because you only care about an arithmetic combination that makes $0$ mod $10$, so proving it for one-digit numbers proves the greater claim.

Without a direct verification, consider all $10^3$ triples of one-digit numbers. If $0$ is in the triple, then multiplying all three is a multiple of $10$.

So now consider all $9^3$ triples from 1--9. If $5$ is in the triple, then either both of the other numbers are odd and you can multiply $5(\text{odd}+\text{odd})$ to get a multiple of $10$; or at least one of the others is even and you can multiply all three to get a multiple of $10$.

So now consider all $8^3$ triples from 1--4,6--9. If any number appear twice in the triple, the difference is $0$, and that difference times the third number is a multiple of $10$.

So now consider all $\binom{8}{3}$ triples from 1--4,6--9 without repetition. If a number and its complement mod $10$ are both in the triple, then sum those and multiply by the third to get a multiple of $10$.

So now consider all $\binom{4}{3}\cdot2^3$ triples from 1--4,6--9 without repetition where no two numbers sum to $10$. We are down to only $32$ triples to consider, and inspecting them directly is not so bad. First, consider the $16$ triples with two or three members from 1--4.

$$\underline{12}3\ \color{magenta}{\underline{1}2\underline{4}}\ \color{orange}{\underline{1}2\underline{6}}\ \color{blue}{127}\ \underline{13}4\ \color{blue}{136}\ 1\underline{38}\ 1\underline{47}\ \color{magenta}{\underline{14}8}\ \color{magenta}{\underline{23}4}\ \color{magenta}{\underline{23}6}\ 2\underline{39}\ \color{orange}{\underline{2}4\underline{7}}\ \color{orange}{2\underline{49}}\ \color{orange}{\underline{3}4\underline{8}}\ 3\underline{49}$$

Blue sum to $0$ mod $10$.

Magenta have an (underlined) pair that sum to $5$ (or $15$) and the third number is even, so that sum times the third number is a multiple of $10$.

Orange have an (underlined) pair whose difference is $5$ and the third number is even, so that difference times the third number is a multiple of $10$.

The remaining black all have an (underlined) pair that sums to the third (mod $10$) so adding that pair and then subtracting the third makes a multiple of $10$.

Note that if we negate all members of one of these triples, we get the triples with two or more in 6--9. The same operations give a multiple of $10$ since we only ever either add and subtract [with the blue and black], or add/subtract and then multiply by an even number [with magenta and orange]. So we've indirectly checked all $32$ of the lingering cases.

• I was hoping to avoid "too many" cases, but this is a pretty modest number of them. And inspecting the final 32 wasn't so bad, at least when they were color-coded :) Would that I could split a bounty, I like all of the answers... – pjs36 Nov 6 '17 at 2:40
• @pjs36 Just FYI, your comment somewhere else got me thinking about how the cases could be cut down to just $16$. – alex.jordan Nov 6 '17 at 3:04

Note: This answer has the focus to reduce the number of variants which are to study by consequently using modular arithmetic.

• $a,b,c\in\mathbb{Z}$

We are looking for multiples of $10$ when doing addition, subtraction and multiplication. Since we have \begin{align*} (a\, \mathrm{mod}\, (10)) + (b\, \mathrm{mod}\, (10)) &\equiv (a+b)\, \mathrm{mod}\, (10) \\ (a\, \mathrm{mod}\, (10)) - (b\, \mathrm{mod}\, (10)) &\equiv (a-b)\, \mathrm{mod}\, (10) \\ (a\, \mathrm{mod}\, (10)) \cdot (b\, \mathrm{mod}\, (10)) &\equiv (a\cdot b)\, \mathrm{mod}\, (10) \\ \end{align*}

it is sufficient to consider $\color{blue}{a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}}$.

• $a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}$

If two of the three numbers are congruent modulo $10$, let's say $a\equiv b\, \mathrm{mod}\, (10)$ we obtain \begin{align*} \color{blue}{(a-b)\cdot c}\equiv 0\cdot c\color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}

and we are done. In the following we may WLOG assume: $a<b<c$

• $a,b,c\in\{0,1,2,3,4,5,6,7,8,9\},a<b<c$

If one of them, let's say $a$ is zero we obtain \begin{align*} \color{blue}{a\cdot b\cdot c}&\equiv 0\cdot b\cdot c \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}

and we are done.

• $a,b,c\in\{1,2,3,4,5,6,7,8,9\}, a<b<c$

If one of them, let's say $a$ is equal to $5$ we consider two cases.

First case: One of $b$ or $c$ is even. Let's assume $b$ is even. It follows \begin{align*} a&\equiv 0\, \mathrm{mod}\, (5)\\ b&\equiv 0\, \mathrm{mod}\, (2)\\ ab&\equiv 0\, \mathrm{mod}\, (10)\\ \end{align*} and we obtain \begin{align*} \color{blue}{a\cdot b\cdot c}&\equiv 0\cdot c \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}

Second case: Both, $b$ and $c$ are odd. It follows from

\begin{align*} b\equiv 1\, \mathrm{mod}\, (2)\\ c\equiv 1\, \mathrm{mod}\, (2)\\ (b+c)\equiv 0\, \mathrm{mod}\, (2)\\ \end{align*}

and we obtain \begin{align*} \color{blue}{a\cdot (b+c)} \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}

• $a,b,c\in\{1,2,3,4,6,7,8,9\}, a<b<c$

Since

\begin{align*} 1\equiv (1-10)\equiv (-9)\, \mathrm{mod}\, (10)\\ 2\equiv (2-10)\equiv (-8)\, \mathrm{mod}\, (10)\\ 3\equiv (3-10)\equiv (-7)\, \mathrm{mod}\, (10)\\ 4\equiv (4-10)\equiv (-6)\, \mathrm{mod}\, (10)\\ \end{align*}

we can restrict the attention to $\color{blue}{a,b,c\in\{1,2,3,4\},a<b<c}$.

• $a,b,c\in\{1,2,3,4\}, a<b<c$

Conclusion: Doing arithmetic with three integers $a,b,c\in\mathbb{Z}$, each of them occurring once and using one or more of the operations addition, subtraction, multiplication and negation ($a \rightarrow -a$) we can always obtain an integer value which is a multiple of $10$.

• This is a nice, thorough answer. Allowing negation is an interesting point. While not explicitly mentioned, I feel like there should be some algebraic trickery that lets us convert negation to addition, subtraction, and multiplication with our same three numbers; e.g., $(-b + a) \cdot c = -(b - a)\cdot c$. Maybe, but maybe it would just serve to introduce more cases, even though it's intuitively clear that negation doesn't really add anything new. – pjs36 Nov 6 '17 at 2:48
• @pjs36: Thanks. :-) Allowing negation is a modest way of cheating somewhat. In fact negation $-a=0-a$ is equivalent to add 0 to $\{+,-,\cdot,(,)\}$. To me it is not so clear that it doesn't introduce anything new. Keep in mind that $-(a-b)$ for instance also uses negation. – Markus Scheuer Nov 6 '17 at 7:16
• @pjs36 I agree, this is where I spend most of the effort in my answer and I would love to see a simpler way to prove that negation can always be eliminated :). – Erick Wong Nov 6 '17 at 18:27
• @MarkusScheuer It’s fine that $-(a-b)$ uses negation: the point of my argument is that all interior negations can be bubbles up to the exterior. Once we have that, then $a-b$ works just as well as $-(a-b)$ for producing zeroes, so we eliminate negation altogether. – Erick Wong Nov 6 '17 at 18:30
• @ErickWong: Ahh, now I see your point! Good argument. Your're right, of course. :-) (+1) – Markus Scheuer Nov 7 '17 at 17:59


Non-commutative or commutative ring or subring with $x^2 = 0$

Does there exist a non-commutative or commutative ring or subring $R$ with $x \cdot x = 0$ where $0$ is the zero element of $R$, $\cdot$ is multiplication secondary binary operation, and $x$ is not zero element, and excluding the case where addition (abelian group operation) and multiplication of two numbers always become zero?

Edit: most seem to be focused on non-commutative case. What about commutative case?

Edit 2: It is fine to relax the restriction to the following: there exists countably (and possibly uncountable) infinite number of $x$'s in $R$ that satisfy $x \cdot x = 0$ (so this means that there may be elements of ring that do not satisfy $x \cdot x =0$) excluding the case where addition and multiplication of two numbers always become zero.

• I forgot to write commutative. So both commutative and non-commutative cases. – user69886 Mar 28 '13 at 13:44
• Regarding your edit: If you want arbitrary elements of this type, just take products or tensor products. So take $k[x_1]/(x_1^2) \times k[x_2]/(x_2^2) \times \dotsc$ or $k[x_1,x_2,\dotsc]/(x_1^2,x_2^2,\dotsc)$. – Martin Brandenburg Mar 28 '13 at 14:29
• @user69886: I updated my answer with a commutative example. In general and for the future, please do not change your question after answers has been given. In general if you have another question, then just post that as a separate question. – Thomas Mar 28 '13 at 14:49

Example for the non-commutative case: $$\pmatrix{0 & 1 \\ 0 & 0} \pmatrix{0 & 1 \\ 0 & 0} = \pmatrix{0 & 0 \\ 0 & 0}.$$ Example for the commutative case: Consider the ring $\mathbb{Z} / 4\mathbb{Z}$. What is $2^2$ in this ring?

• See my answer for a conceptual view of the genesis of this example. – Math Gems Mar 28 '13 at 15:54

Sure, take $k$ any ring and $R=k[x]/(x^2)$. As suggested by Martin Brandenburg, a good example is $k=\mathbb R$, in which case $R$ becomes the ring of dual numbers. The wikipedia site lists a lot of properties, which might be interesting.

• Or almost the same, $\mathbb{Z}/4$. – Martin Brandenburg Mar 28 '13 at 13:40
• It's commutative. Though if we take a noncommutative base ring instead of $k$.. – Berci Mar 28 '13 at 13:41
• @Berci: Whoops, I misread that. – Jesko Hüttenhain Mar 28 '13 at 13:42
• I forgot to write "or commutative.." – user69886 Mar 28 '13 at 13:44
• Jesko, you may perhaps add a link to en.wikipedia.org/wiki/Dual_number – Martin Brandenburg Mar 28 '13 at 13:53

It is easy if you know about quotient rings, e.g. $\rm\:\Bbb Z/n^2,\ \Bbb Z[x]/(x^2).\:$ If you don't know about quotient rings, is there still a simple example? Yes! One quickly checks that $\rm\,\Bbb Z^2$ is a ring under pointwise addition and multiplication and, that the set of linear maps on $\rm\,\Bbb Z^2\,$ form a ring under addition and map composition. Now note that the (right) shift map $\rm\: S(a,b) = (0,a)\:$ is linear, and $\rm\,S^2 = 0\,$ since $\rm\: S^2(a,b) = S(0,a) = (0,0).\:$

Remark  If you know matrix rings, note that the matrix of the shift map is that in Thomas's answer. But knowledge of matrices is not required to understand the above simple example.

Where does this example come from? It arises from a general construction. It is, in fact, a linear representation of the generic example $\rm\:R[x]/(x^2),\:$ whose additive group is $\rm\:R^2,\:$ where $\rm\:x\:$ acts as a (right) shift map $\rm\:x(a+bx) = 0+ax,\:$ i.e. $\rm\:(a,b)\mapsto (0,a).\:$ Similarly any ring $\rm\,R\,$ has such a linear representation as a subring of the ring of linear maps on its underling additive group - the so-called (left) regular representation of $\rm\,R.\:$ This is a ring-theoretic analogue of Cayley's theorem, that a group $\rm\,G\,$ may be represented as a subgroup of the group of bijections (permutations) on $\rm\,G.\:$ In both cases the representation arises by sending each element $\rm\:r\:$ into its associated (left) multiplication map $\rm\: x\:\to\:r\:x\:.\:$ See this MO question for further discussion.

As another important example, this view provides a natural way to construct the polynomial ring $\rm\,R[x]\,$ as a subring of the ring of linear maps on $\rm\,R^\Bbb N = (f_0,f_1,f_2,\ldots)\:$ generated by $\rm\,R\,$ and $\rm\,x\,$ i.e. by all constants $\rm\:(r,0,0,\ldots)$ and by the shift map $\rm\,x : (f_0,f_1,f_2,\ldots)\mapsto (0,f_0,f_1,\ldots).$ See also the presentation in this post.

It is worth emphasis that the ring $\rm\:R[x]/(x^2)\:$ is known as the algebra of dual numbers over $\rm\,R,\,$ and that it frequently proves quite handy in various contexts, e.g. when studying (higher) derivatives, or tangent/jet spaces, and when transferring properties of homomorphisms to derivations

• Actually $S=x$ on $\mathbb{Z}[x]/(x^2)$. So these are the same examples in disguise. – Martin Brandenburg Mar 28 '13 at 14:17
• @martin Yes, I meant to segue into that - see the added remark. – Math Gems Mar 28 '13 at 14:57

Yes, for example consider the noncommutative polynomial ring $\Bbb R\langle x,y\rangle$ (so here $xy\ne yx$), and let $R$ be its quotient by the ideal $(x^2)$ generated by $x^2$.

• Non-commutative polynomial rings are often called free (associative) algebras and denoted by other brackets, namely $R\langle x_1,\dotsc,x_n\rangle$. – Martin Brandenburg Mar 28 '13 at 13:48


# Tossing a matchstick randomly onto an infinite grid

I was given this problem recently in a job interview in which I did not succeed, however I found the problem itself very interesting, and have been trying to solve it since.

Question:

Given matchstick of length 1, and an infinite grid of similar matchsticks (i.e. an infinite grid of 1x1 squares), if you toss the matchstick randomly onto the grid, what is the probability that the matchstick lands overlapping at least one of the matchsticks in the grid?

My approach:

I decided to first define 3 variables:

• $x =$ horizontal distance between lower end of landing matchstick and nearest grid matchstick to the left
• $y =$ vertical distance between lower end of landing matchstick and nearest grid matchstick below
• $\theta$ = landing angle of matchstick with grid horizontal

By definition:

• $x, y$ ~ $U(0, 1)$
• $\theta$ ~ $U(0, \pi)$
• $x, y, \theta$ are all independent of each other

I then split the events of overlapping the horizontal and the vertical matchsticks in the grid, such that:

$$P(Overlap) = P(OverlapsVertical) + P(OverlapsHorizontal) - P(OverlapsVertical \bigcup OverlapsHorizontal)$$

• $P(OverlapsVertical) = P(y + sin(\theta) >= 1)$
• $P(OverlapsHorizontal) = P(x + cos(\theta) >= 1)$

For overlapping vertical, because both y and theta are uniformly distributed, I could calcluate $P(OverlapsVertical)$ by:

• Constructing the rectangle of possible values for $y$ and $\theta$
• Integrating to find the proportion of area of this rectangle that satisfied the condition $y + sin(\theta) >= 1$

Doing the same for the horizontal case, I found that $P(OverlapsVertical) = P(OverlapsHorizontal) = 2/\pi$.

However, I realised that the events $OverlapsVertical$ and $OverlapsHorizontal$ are not independent, because they are both dependent on $\theta$, specifically $sin(\theta)$ and $cos(\theta)$ which are inversely related, and therefore:

• $P(OverlapsVertical \bigcup OverlapsHorizontal) \not= P(OverlapsVertical) * P(OverlapsHorizontal)$

I then became stuck, and wondered if my approach to the entire problem was wrong. Is there a better way to solve this problem? Any input greatly appreciated. Thanks!

• – Bram28 Jan 2 '18 at 22:07
• @Bram28 thanks for the quick reply. I see how that applies to the individual events of crossing the vertical and crossing the horizontal (my problem is specifically Buffon's needle where t = l = 1, and hence agrees with my answer 2/π), however as far as I can tell it does not answer the question of crossing either the vertical or the horizontal, where these two events are not independent nor mutually exclusive. – Rory Devitt Jan 2 '18 at 22:20
• Out of curiosity, which position were you applying for ? – Gabriel Romon Jan 2 '18 at 22:24
• @GabrielRomon it was a quantitative trading position at a hedge fund – Rory Devitt Jan 2 '18 at 22:25
• And the interviewer didn't give you a hint or a nudge in some direction ? – Gabriel Romon Jan 2 '18 at 22:31

The if the midpoint lands inside a box of dimensions: $(1-\cos\theta)\times(1-\sin\theta),$ then then the match will fit inside that square.
Due to the symmetry of the grid we only need to analyze $\theta \in [0, \frac {\pi}{4}]$
$\frac 1{\frac {\pi}{4}}\int_0^\frac {\pi}{4} (1-\cos\theta)\cdot(1-\sin\theta)\ d\theta$