It is currently 27 Jun 2017, 19:33

### GMAT Club Daily Prep

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

Customized
for You

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

Track

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

Practice
Pays

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Why this is illegal?

Author Message
TAGS:

### Hide Tags

Manager
Joined: 12 Dec 2012
Posts: 156
Location: Poland

### Show Tags

18 Jan 2013, 14:40
1
KUDOS
1
This post was
BOOKMARKED
Could someone please explain me why this is illegal?:

$$\sqrt{x}+\sqrt{y}>0$$

$$\sqrt{x}>-\sqrt{y}$$

$$(\sqrt{x})^2>(-\sqrt{y})^2$$

$$x>y$$

While this is legal:

$$\sqrt{x}-\sqrt{y}>0$$

$$\sqrt{x}>\sqrt{y}$$

$$(\sqrt{x})^2>(\sqrt{y})^2$$

$$x>y$$

Thank you.
_________________

If I answered your question with this post, use the motivating power of kudos!

Math Expert
Joined: 02 Sep 2009
Posts: 39723
Re: Why this is illegal? [#permalink]

### Show Tags

18 Jan 2013, 15:03
4
KUDOS
Expert's post
1
This post was
BOOKMARKED
HumptyDumpty wrote:
Could someone please explain me why this is illegal?:

$$\sqrt{x}+\sqrt{y}>0$$

$$\sqrt{x}>-\sqrt{y}$$

$$(\sqrt{x})^2>(-\sqrt{y})^2$$

$$x>y$$

While this is legal:

$$\sqrt{x}-\sqrt{y}>0$$

$$\sqrt{x}>\sqrt{y}$$

$$(\sqrt{x})^2>(\sqrt{y})^2$$

$$x>y$$

Thank you.

Writing $$x>y$$ from $$\sqrt{x}+\sqrt{y}>0$$ is not right. We have that the sum of two non-negative values ($$\sqrt{x}$$ and $$\sqrt{y}$$) is greater than zero. How can we know that $$x>y$$ from that? With the same logic you could get that $$y>x$$. Right?

Algebraic explanation: we can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;
But consider the case when one side is negative: $$-2<1$$ --> if we square we get $$4<1$$, which is not right. So, squaring an inequality where one side is negative won't always give the correct result.

In the first case $$-\sqrt{y}\leq{0}$$, thus we cannot apply squaring. While in the second case both parts of the inequality are non-negatve, thus we can safely square.

GENERAL RULE:
A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

Hope it helps.

P.S. Can you please post the question from which you took that example? Thank you.
_________________
Manager
Joined: 12 Dec 2012
Posts: 156
Location: Poland
Re: Why this is illegal? [#permalink]

### Show Tags

18 Jan 2013, 16:09
1
KUDOS
It helped a lot, thanks! The problem posted in DS Section:

a-x-y-and-b-x-y-if-a-2-b-2-what-is-the-value-of-y-146004.html
_________________

If I answered your question with this post, use the motivating power of kudos!

Math Expert
Joined: 02 Sep 2009
Posts: 39723
Re: Why this is illegal? [#permalink]

### Show Tags

18 Jan 2013, 16:20
HumptyDumpty wrote:
It helped a lot, thanks! The problem posted in DS Section:

a-x-y-and-b-x-y-if-a-2-b-2-what-is-the-value-of-y-146004.html

Just posted a solution: a-x-y-and-b-x-y-if-a-2-b-2-what-is-the-value-of-y-146004.html#p1170874

By the way, we can solve this question without squaring inequalities.
_________________
Manager
Joined: 12 Dec 2012
Posts: 156
Location: Poland
Re: Why this is illegal? [#permalink]

### Show Tags

18 Jan 2013, 16:50
Bunuel wrote:

Just posted a solution: a-x-y-and-b-x-y-if-a-2-b-2-what-is-the-value-of-y-146004.html#p1170874

By the way, we can solve this question without squaring inequalities.

Yes that I know, but I had that gap which you've just dismissed .
_________________

If I answered your question with this post, use the motivating power of kudos!

Manager
Joined: 28 Dec 2012
Posts: 111
Location: India
Concentration: Strategy, Finance
WE: Engineering (Energy and Utilities)
Re: Why this is illegal? [#permalink]

### Show Tags

31 Jan 2013, 01:44
sqrt x + sqrt y = 0 fine

sqrt can only be positive

sum of these 2 positive numbers can be 0 only if each is equal to zero so only one possibility.

sqrt x = - sqrt y LHS positive, RHS negative.... ????? again, each side should be 0

Now when u r trying to square each side,

u r multiply with a positive quantity on LHS and negative quantity on RHS

Inequalities should be either multiplied by a positive number or a negative number on each side, with sign changing in the latter case.

Thanks.

KUDOS if u appreciate
_________________

Impossibility is a relative concept!!

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16014
Re: Why this is illegal? [#permalink]

### Show Tags

20 Nov 2014, 06:05
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Why this is illegal?   [#permalink] 20 Nov 2014, 06:05
Similar topics Replies Last post
Similar
Topics:
Why am I so stupid? 4 26 Mar 2016, 14:43
2 Why multiply by 1.5? 3 24 Nov 2014, 10:55
1 A quick question on fractions: why THIS and not THAT? 2 06 Jul 2013, 04:52
why can't I divide by |y| ? 4 03 Nov 2011, 13:48
Why is Algebra so hard for me 6 03 Feb 2011, 08:59
Display posts from previous: Sort by