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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