1. ## This calculus lark has gone right over my head!

This is the most recent problem I can't do!

State the contrapositive and give a proof of the contropositive:

If 1/2(a^2 + b^2) >ab then a/=/b (a not equal to b)

I guess the contrapositve is If Not a/=/b then Not 1/2(a^2 + b^2) > ab

But now what?!

2. Can someone please go over what they'd do for this answer, possibly even the full answer so I can go through it, learn how it's to be done then move on with the next 5 questions I have!

Chris

3. Originally Posted by chr91
This is the most recent problem I can't do!

State the contrapositive and give a proof of the contropositive:

If 1/2(a^2 + b^2) >ab then a/=/b (a not equal to b)

I guess the contrapositve is If Not a/=/b then Not 1/2(a^2 + b^2) > ab

But now what?!
$\displaystyle \frac12(a^2+b^2)>ab~\implies~a^2+b^2>2ab~\implies~ a^2-2ab+b^2>0$

$\displaystyle (a-b)^2>0$ is true if $\displaystyle a-b\neq 0$

And now state the contrapositive more precisely.

4. Originally Posted by earboth
$\displaystyle \frac12(a^2+b^2)>ab~\implies~a^2+b^2>2ab~\implies~ a^2-2ab+b^2>0$

$\displaystyle (a-b)^2>0$ is true if $\displaystyle a-b\neq 0$

And now state the contrapositive more precisely.
Thanks.

Well now we have (a-b)^2 > 0 implies a/=/b

Taking the contropositive we get Not a/=/b implies not (a-b)^2 > 0

a=b implies (a-b)^2 < or the same as 0

If a=b, (a-b)^2 = 0 therefore the contropositive has been proved.

Have I done that correctly?