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Why this is illegal?

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Why this is illegal? [#permalink] New post 18 Jan 2013, 13:40
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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.
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Re: Why this is illegal? [#permalink] New post 18 Jan 2013, 14:03
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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.
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Re: Why this is illegal? [#permalink] New post 18 Jan 2013, 15:09
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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
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Re: Why this is illegal? [#permalink] New post 18 Jan 2013, 15:20
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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.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Why this is illegal? [#permalink] New post 18 Jan 2013, 15:50
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 :).
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Re: Why this is illegal? [#permalink] New post 31 Jan 2013, 00:44
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.


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Re: Why this is illegal? [#permalink] New post 20 Nov 2014, 05:05
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Re: Why this is illegal?   [#permalink] 20 Nov 2014, 05:05
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