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If x and y are integers, is x > y?

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Re: If x and y are integers, is x > y?  [#permalink]

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New post 07 Jun 2018, 19:54
I'll try to give the simplest solution I could think of:

---------------------------------------------------------------------------

1. \(x + y > 0\)

Therefore: \(x > -y\)

Example: if x = 3 and y = -2 , this is satisfied. This also satisfied x > y (that is, 3 > -2.)
Counterexample: if x = 1 and y = 1, this is satisfied. However, this violates the x > y equation.
We have found both an example and a counterexample - therefore, 1 by itself is insufficient.

---------------------------------------------------------------------------

2. \(y^x < 0\)

The only way this is true is if x is odd and y is negative.

But there's an infinite number of possibilities where x is odd and y is negative. Therefore, 2 by itself is insufficient.

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What about equation 1 and equation 2 combined?

If we know y is negative, then the equation in 1:

\(x > -y\)

becomes

\(x > y\) (does that look familiar?)

Therefore, by definition, both 1 and 2 together are necessary to solve the problem.
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Re: If x and y are integers, is x > y?  [#permalink]

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New post 22 Aug 2018, 10:05
Walkabout wrote:
If x and y are integers, is x > y?

(1) x + y > 0
(2) y^x < 0



combining two statements Y must be negative and \(x + y > 0\) means that \(y^x < 0\) could be following values \(-3^5\), \(-3^9\) etc X will always be greater
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Re: If x and y are integers, is x > y?  [#permalink]

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New post 22 Aug 2018, 14:52
The interesting issue here may be explicitly presented below:

\({y^x} < 0\,\,\,;\,\,\,x + y > 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x > y\)

To prove that, let´s suppose, on the contrary, that we could have SIMULTANEOUSLY the following:

(1) \({y^x} < 0\)
(2) \(x + y > 0\)
(3) \(x \leqslant y\)

The contradiction follows:
\(\left( 1 \right)\,\,\,\, \Rightarrow \,\,\,\,\,y < 0\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( 3 \right)} \,\,\,x < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,x + y < 0\,\,\,\,\, \Rightarrow \,\,\,\,\left( 2 \right)\,\,{\text{contradicted}}\)
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Re: If x and y are integers, is x > y?  [#permalink]

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New post 16 Sep 2018, 11:12
Hi,
here are my two cents for this questions

Question is IS x>y
or we can rephrase this as IS x-y>0

Lets understand in which conditions is x-y>0

x>0,y>0, and |x|>|y|
x<0,y<0, and |y|>|x|
x>0,y<0, and |x|>|y|
x<0,y>0, this case is not possible

So Statement A:
x+y>0
we can see following cases
x>0, y>0
x<0, y>0 and |y|>|x|
x>0,y<0 and |x|>|y|
x<0,y<0 this case is not possible

We can't conclude anything if x>y
Hence A is insufficient


From Statement B:
\(y^x\)<0
so we have two cases
y is negative and x is positive and odd
y is negative and x is negative and odd

So we can't conclude anything from above if x>y
Hence B is insufficient

Now combining the above two we have

there is only one conditions that satisfies both Statement A & B
which is
x>0 and y<0.
So we can conclude x>y
Hence C is sufficient.
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Re: If x and y are integers, is x > y? &nbs [#permalink] 16 Sep 2018, 11:12

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