Finding counter-examples

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

franzliszt

23531 posts

franzliszt

#1 h Feb 5, 2021, 5:49 PM • 2 Y

Y by mathematicsy, A4r43r457yhyt

Under what conditions is $\sqrt{a^2+b^2}=a+b$ true, where $a$ and $b$ are real numbers?

$\textbf{(A) }$ It is never true.
$\textbf{(B) }$ It is true if and only if $ab=0$ .
$\textbf{(C) }$ It is true if and only if $a+b\ge 0$ .
$\textbf{(D) }$ It is true if and only if $ab=0$ and $a+b\ge 0$ .
$\textbf{(E) }$ It is always true.

This post has been edited 1 time. Last edited by franzliszt, Feb 5, 2021, 6:54 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

akid

653 posts

akid

#2 h Feb 5, 2021, 5:49 PM

Y by

My memory is horrible somehow

This post has been edited 1 time. Last edited by akid, Feb 5, 2021, 5:55 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

rickthekidfan13

24 posts

rickthekidfan13

#3 h Feb 5, 2021, 5:50 PM • 1 Y

Y by Imayormaynotknowcalculus

The answer is D

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ultroid999OCPN

1802 posts

Ultroid999OCPN

#4 h Feb 5, 2021, 5:50 PM • 1 Y

Y by sugar_rush

lol got baited and picked B (correct answer is D)
rip easy 6 points and AIME qual

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

wolfpack

1274 posts

wolfpack

#5 h Feb 5, 2021, 6:07 PM

Y by

Full Solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ShockFish

2394 posts

ShockFish

#6 h Feb 5, 2021, 6:07 PM

Y by

Why is almost everything in the global feed contests and programs???

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ha_ha_ha

1139 posts

Ha_ha_ha

#7 h Feb 5, 2021, 6:08 PM

Y by

Lol got D after some thought

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

iiRishabii

1155 posts

iiRishabii

#8 h Feb 5, 2021, 6:33 PM

Y by

Squaring both sides, we have

$a^2+b^2=a^2+2ab+b^2.$

Some terms cancel out, leaving us with

$2ab=0.$
$ab=0.$

Also, by the Trivial Inequality, $a+b$ cannot be less than $0$ since the value of $\sqrt{a^2+b^2}$ is always greater than or equal to $0.$ Therefore, the answer is (D.)

This post has been edited 1 time. Last edited by iiRishabii, Feb 5, 2021, 6:33 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Kinglogic

98 posts

Kinglogic

#9 h Feb 5, 2021, 6:34 PM

Y by

yeah it's D. If a or b is negative then the LHS will be positive but RHS is negative. So it's D instead of B.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

franzliszt

23531 posts

franzliszt

#10 h Feb 5, 2021, 6:50 PM

Y by

ShockFish wrote:

Why is almost everything in the global feed contests and programs???

AMCs were yesterday.

Regarding the main problem, it was fairly straightforward. If unsure, find counter examples.

$A$ is false: $(a,b)=(0,0)$ .
$B$ is false: $(a,b)=(-1,0)$
$C$ is false: $(a,b)=(-1,1)$
$E$ is false: $(a,b)=(-1,1)$

So pick $D$ .

This post has been edited 1 time. Last edited by franzliszt, Feb 6, 2021, 2:09 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

sugar_rush

1341 posts

sugar_rush

#11 h Feb 8, 2021, 7:58 PM

Y by

Isn't this a repeat of AMC 10 2006A #11?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ilovepizza2020

12154 posts

ilovepizza2020

#12 h Feb 8, 2021, 8:08 PM • 1 Y

Y by tenebrine

sugar_rush wrote:

Isn't this a repeat of AMC 10 2006A #11?

No, the square root in this problem makes it different.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

sugar_rush

1341 posts

sugar_rush

#13 h Feb 8, 2021, 8:09 PM

Y by

ilovepizza2020 wrote:

sugar_rush wrote:

Isn't this a repeat of AMC 10 2006A #11?

No, the square root in this problem makes it different.

@above why?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

peace09

5419 posts

peace09

#14 h Feb 8, 2021, 8:14 PM

Y by

Obviously, $a^2+b^2$ is positive. Then, the square root of a positive real number is also a positive real number. In other words, if $a$ is a positive real constant, then the equation $x=\sqrt{a}$ has one solution and the equation $x^2=a$ has two solutions.

Also note that if $f(x)=\sqrt{x}$ is a function with domain $[0, \infty),$ then there must be one output for every input.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ilovepizza2020

12154 posts

ilovepizza2020

#15 h Feb 8, 2021, 8:14 PM • 1 Y

Y by tenebrine

sugar_rush wrote:

ilovepizza2020 wrote:

sugar_rush wrote:

Isn't this a repeat of AMC 10 2006A #11?

No, the square root in this problem makes it different.

@above why?

This problem has to have $a+b\geq 0$ , while the other doesn't.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

sugar_rush

1341 posts

sugar_rush

#16 h Feb 8, 2021, 8:22 PM • 1 Y

Y by Mango247

@above you're incorrect

Let $a=-1, b=0$
Then $a^{2}+b^{2}=1$
If someone asked you to find "the square root of $1$ ", then there are two possibilities: $-1$ or $1$ .
Since $-1=a+b$ , B should be fine?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

peace09

5419 posts

peace09

#17 h Feb 8, 2021, 8:23 PM

Y by

Answer this question:

Is $f(x)=\sqrt{x}$ a function?

There is a difference between $y=\sqrt{1}$ and $y^2=1.$

This post has been edited 1 time. Last edited by peace09, Feb 8, 2021, 8:25 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ReaperGod

1579 posts

ReaperGod

#18 h Feb 8, 2021, 8:28 PM

Y by

sugar_rush wrote:

@above you're incorrect

Let $a=-1, b=0$
Then $a^{2}+b^{2}=1$
If someone asked you to find "the square root of $1$ ", then there are two possibilities: $-1$ or $1$ .
Since $-1=a+b$ , B should be fine?

if it uses the symbol, only 1 answer

if it says it with words, then 2 answers unless it is 0

@below exactly

This post has been edited 1 time. Last edited by ReaperGod, Feb 8, 2021, 8:52 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

peace09

5419 posts

peace09

#19 h Feb 8, 2021, 8:29 PM • 1 Y

Y by Mango247

@#18: Not sure what you mean. You're saying there's a difference between The square root of $x$ and $\sqrt{x}?$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ilovepizza2020

12154 posts

ilovepizza2020

#20 h Feb 8, 2021, 9:43 PM • 1 Y

Y by tenebrine

peace09 wrote:

@#18: Not sure what you mean. You're saying there's a difference between The square root of $x$ and $\sqrt{x}?$

I don't think so. The main difference we are trying to stress is that $x^2=a$ is different from $x=\sqrt{a}$ , and not that words are different from numbers.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

pooh_bear

756 posts

pooh_bear

#21 h Feb 8, 2021, 9:54 PM

Y by

sugar_rush wrote:

@above you're incorrect

Let $a=-1, b=0$
Then $a^{2}+b^{2}=1$
If someone asked you to find "the square root of $1$ ", then there are two possibilities: $-1$ or $1$ .
Since $-1=a+b$ , B should be fine?

By the definition of the square root, $\sqrt{x^2} = |x|$ . Which makes $B$ incorrect.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

donian9265

541 posts

donian9265

#23 h Feb 8, 2021, 10:03 PM

Y by

^above
Wait, do you mean D?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

CoolCarsOnTheRun

2846 posts

CoolCarsOnTheRun

#24 h Feb 8, 2021, 10:07 PM

Y by

I didn't read the post, but as the correct answer is D and not B, I assume they mean B and not D (or what they said).

This post has been edited 1 time. Last edited by CoolCarsOnTheRun, Feb 8, 2021, 10:07 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ReaperGod

1579 posts

ReaperGod

#25 h Feb 8, 2021, 10:22 PM

Y by

ilovepizza2020 wrote:

peace09 wrote:

@#18: Not sure what you mean. You're saying there's a difference between The square root of $x$ and $\sqrt{x}?$

I don't think so. The main difference we are trying to stress is that $x^2=a$ is different from $x=\sqrt{a}$ , and not that words are different from numbers.

actually they are different. the words are talking about the first one

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ilovepizza2020

12154 posts

ilovepizza2020

#26 h Feb 8, 2021, 11:06 PM • 4 Y

Y by tenebrine, Mango247, Mango247, Mango247

ReaperGod wrote:

ilovepizza2020 wrote:

peace09 wrote:

@#18: Not sure what you mean. You're saying there's a difference between The square root of $x$ and $\sqrt{x}?$

I don't think so. The main difference we are trying to stress is that $x^2=a$ is different from $x=\sqrt{a}$ , and not that words are different from numbers.

actually they are different. the words are talking about the first one

Wdym? The square root of 9 is 3, not -3.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

channing421

1353 posts

channing421

#27 h Feb 8, 2021, 11:31 PM • 1 Y

Y by megarnie

sugar_rush wrote:

@above you're incorrect

Let $a=-1, b=0$
Then $a^{2}+b^{2}=1$
If someone asked you to find "the square root of $1$ ", then there are two possibilities: $-1$ or $1$ .
Since $-1=a+b$ , B should be fine?

I don't know where u got this idea from, but square roots are always positive. For example, $\sqrt9=3$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

peace09

5419 posts

peace09

#28 h Feb 8, 2021, 11:32 PM

Y by

TL;DR

$\sqrt{x^2}=|x|.$ For positive $x,$ $\sqrt{x}$ is a constant, meaning it is one number. While the equation $x=\sqrt{a}$ has ONE root, the equation $x^2=a$ has TWO roots. These two equations share a root in common, namely, $\sqrt{a}.$ The root of the second equation that is not a root of the first equation is $-\sqrt{a}.$

If $\sqrt{a}$ were equal to two numbers, namely, the two roots for $x^2=a,$ then the function $f(a)=\sqrt{a}$ wouldn't be a function, which is clearly not true.

@#27: Agreed.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

skrublord420

1277 posts

skrublord420

#29 h Feb 9, 2021, 10:19 PM • 6 Y

Y by peace09, Toinfinity, dchen, mira74, sub_math, Taco12

Perhaps this video will help settle the debate.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

sugar_rush

1341 posts

sugar_rush

#31 h Jul 1, 2021, 1:14 AM • 2 Y

Y by samrocksnature, mira74

The linchpin that cost me AIME qualification from 12A.

solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

samrocksnature

8791 posts

samrocksnature

#32 h Jul 1, 2021, 1:28 AM

Y by

sugar_rush wrote:

The linchpin that cost me AIME qualification from 12A.

solution

:/

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Supernova283

1296 posts

Supernova283

#33 h Jul 1, 2021, 3:29 AM

Y by

sugar_rush wrote:

@above you're incorrect

Let $a=-1, b=0$
Then $a^{2}+b^{2}=1$
If someone asked you to find "the square root of $1$ ", then there are two possibilities: $-1$ or $1$ .
Since $-1=a+b$ , B should be fine?

Actually, the square root of $1$ is $1$ , even though both $-1^2$ and $1^2$ equal $1$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

rama1728

800 posts

rama1728

#34 h Jul 1, 2021, 5:07 AM • 1 Y

Y by sugar_rush

Look guys, if the square root of 1 is both 1 and -1, then 1=-1, and this leads to different paradoxes. To avoid this nasty war, we take just one value. Now, if that one value was negative, then $\sqrt{1}=-1$ , $\sqrt{1}=\sqrt{1}\cdot\sqrt{1}=-1\cdot -1=1$ , a contradiction. Therefore, we take the positive value of the square root, and I hope there are no problems in my argument.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Archeon

5970 posts

Archeon

#35 h Jul 1, 2021, 4:23 PM • 1 Y

Y by centslordm

By convention, the $\sqrt{\cdot}$ symbol is used to represent the principal square root of its input.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

megarnie

5611 posts

megarnie

#36 h Oct 6, 2021, 11:28 PM

Y by

Solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Inventor6

558 posts

Inventor6

#37 h Oct 8, 2021, 5:33 AM

Y by

We square both sides and we get
a^2 + b^2 = a^2 + b^2 + 2ab
so
0 = 2ab

Also, if a + b is negative, then the square root of a^2 + b^2 would be negative, but we have to use the positive square root.

The answer is D

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mm999aops

2635 posts

mm999aops

#38 h Oct 8, 2021, 5:56 AM

Y by

imagine not reading all of the choices before answering this lol

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Eveoverrr

9 posts

Eveoverrr

#39 h Nov 7, 2021, 5:59 AM

Y by

I will choose D

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

solasky

1566 posts

solasky

#40 h Nov 7, 2021, 6:19 AM

Y by

sugar_rush wrote:

@above you're incorrect

Let $a=-1, b=0$
Then $a^{2}+b^{2}=1$
If someone asked you to find "the square root of $1$ ", then there are two possibilities: $-1$ or $1$ .
Since $-1=a+b$ , B should be fine?

No, the square root of $1$ has only one value which is $1.$

Sometimes in Algebra 1, people are taught to simplify
$\begin{align*} x^2 &= k \\ x &= \pm \sqrt{k} \end{align*}$ causing the confusion that the square root of $k$ is $\pm \sqrt{k}.$ However, you actually skipped a step, the correct steps will be
$\begin{align*} x^2 &= k \\ |x| &= \sqrt{k} \\ x &= \pm \sqrt{k} \end{align*}$ because $\sqrt{x^2}=|x|.$

If you're still not convinced, just test some values. If $a=0$ and $b=-1,$ $\text{ (B) }$ states that the statement should be true. However $\sqrt{0^2+(-1)^2}=\sqrt{0+1}=1 \ne 0 + (-1) = -1.$

Supernova283 wrote:

sugar_rush wrote:

@above you're incorrect

Let $a=-1, b=0$
Then $a^{2}+b^{2}=1$
If someone asked you to find "the square root of $1$ ", then there are two possibilities: $-1$ or $1$ .
Since $-1=a+b$ , B should be fine?

Actually, the square root of $1$ is $1$ , even though both $-1^2$ and $1^2$ equal $1$ .

Actually, $-1^2=-1.$ I assume that's just a typo and you meant $(-1)^2.$

This post has been edited 2 times. Last edited by solasky, Nov 7, 2021, 6:20 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

NTistrulove

183 posts

NTistrulove

#41 h Dec 20, 2021, 8:27 AM

Y by

franzliszt wrote:

Under what conditions is $\sqrt{a^2+b^2}=a+b$ true, where $a$ and $b$ are real numbers?

$\textbf{(A) }$ It is never true.
$\textbf{(B) }$ It is true if and only if $ab=0$ .
$\textbf{(C) }$ It is true if and only if $a+b\ge 0$ .
$\textbf{(D) }$ It is true if and only if $ab=0$ and $a+b\ge 0$ .
$\textbf{(E) }$ It is always true.

Now first of all I thought of $B$ , but then I saw that square root (-_-).

The answer is actually $D$ , as if $a+b=\sqrt{a^2+b^2}$ then $a+b\geq 0$ and,
$a^2+b^2+2ab=a^2+b^2\implies ab=0$ ...