Bigger Squared Is Still Bigger?

Algebra Level 1

True or False?

For real numbers $x$ and $y$ ,

$x \geq y \quad \Rightarrow \quad x^2 \geq y^2.$

Inspiration .

True False

5 solutions

Aditya Narayan Sharma
Oct 30, 2016

Just the case where it wouldn't fit is for negative numbers. Suppose $x=-k$ then $k\le -y \implies k^2\le y^2 \implies x^2\le y^2$

The actual statement is $|x|\ge |y|\implies x^2\ge y^2$

I've made that mistake so many times! How can we avoid it?

Chung Kevin - 4 years, 7 months ago

Ayush G Rai
Oct 29, 2016

If suppose $3\geq -5.$ Then squaring them would give $9\geq 25$ which is not true.So to generalize,the statement can be false when $y$ is a negative real number whose absolute value is greater than any real number $x.$

To get a big picture, I would draw the function $f(x)=x^2$ . Pick two numbers, and see if their corresponding value matches the inequality. Then you would see that this is only true when both $x$ and $y$ are positive.

Christopher Boo - 4 years, 7 months ago

Roy Bertoldo
Dec 7, 2016

Let X be a positive number and Y be a negative number.

Such as X=2 , Y=-3

Substituting for X and Y:

2 > (-3) but 2^2 = 4 and (-3)^2 = 9 giving a case where X > Y but X^2 < Y^2

Oli Hohman
Dec 6, 2016

I proved this is false by considering the contrapositive of the statement, which is logically equivalent to the original conditional statement.

Contrapositive: If x^2<y^2, then x<y.

Counterexample: x=-1/2 and y=-1. x^2 = (-1/2)^2 = 1/4 y^2 = (-1)^2 = 1

But, -1/2 > -1, so x>y, which contradicts the conclusion of the contrapositive.

That's nice! In fact, finding a counter example for the contrapositive is the same as finding an example for the statement.

Calvin Lin Staff - 4 years, 6 months ago

Kaushik Chandra
Nov 3, 2016

Let's take x=3&y=-5 then x>y while x^2=9 & y^2=25 here y>x

Bigger Squared Is Still Bigger?

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