whether the statement is true or false or sometimes true

[ticka05]

whether the statement is true,false, or sometimes true. provide an example to justify your answer
if both Xand Y are integers and X>Y, then X2>Y2

 

[Loren]

whether the statement is true,false, or sometimes true. provide an example to justify your answer
if both Xand Y are integers and X>Y, then X2>Y2 <<< Need to clarify. Does X2 stand for x[sup:26s9p8ua]2[/sup:26s9p8ua], x[sub:26s9p8ua]2[/sub:26s9p8ua] or x times 2 or something else?

 

[Aladdin]

Let x = 3 & let y = 2

\(\displaystyle x^2=(3)^2=9\)

\(\displaystyle y^2=(2)^2=4\)

x > y --- x^2 > y^2

 

[soroban]

Hello, ticka05!

Whether the statement is true,false, or sometimes true.
Provide an example to justify your answer

if both \(\displaystyle x\) and \(\displaystyle y\) are integers and \(\displaystyle x > y\), then: .\(\displaystyle x^2 > y^2\)

This statement is only sometimes true . . .


Counter-example:

We have: .\(\displaystyle x \:=\:-2,\;y \:=\:-3 \quad\Rightarrow\quad x \:>\:y\)

But \(\displaystyle x^2\) is not greater than \(\displaystyle y^2.\)

. . \(\displaystyle \begin{array}{ccc}x^2 \:=\\text{-}2)^2 \:=\:4 \\ y^2 \:=\\text{-}3)^2 \:=\:9 \end{array}\quad\Rightarrow\quad 4 \:<\:9 \quad\Rightarrow\quad x^2 \:<\:y^2\)

 

[stapel]

Let x = 3 & let y = 2

Examples do not prove, though counter-examples (as provided by a helper) can dis-prove. If you think that this (false) statement is true, you need to do a proof "in full generality":

Suppose that, if x > y, then x[sup:2q8d91vb]2[/sup:2q8d91vb] > y[/sup]. Then:

x[sup:2q8d91vb]2[/sup:2q8d91vb] - y[sup:2q8d91vb]2[/sup:2q8d91vb] > 0

(x - y)(x + y) > 0

x - y > 0 and x + y > 0
or
x - y < 0 and x + y < 0

Then:

x > y and x > -y
or
x < y and x < -y

But we have assumed that x > y, so the second solution set will not work. Looking at the first solution set, we get x > y and x > -y, which means that x > |y|. But we can have x > y without having x > |y|, and we only assumed the former.

This disproves the "suppose" statement, and tells you precisely where it will fail: for x > y but x < |y|.

Once you move into college math and higher, you will learn more about the logic of math proofs, so the reasoning will make more sense. :wink:

The counter-example, already provided, should already make sense. In fact, you can use the dis-proof above to create any number of counter-examples. :lol: