# Prove a^2≥0 always

1. Apr 16, 2012

### jaqueh

1. The problem statement, all variables and given/known data
prove a^2 is positive

I have tried this, but i dont know if it's conclusive:

i) Suppose a≥0
then a+1≥1
then a(a+1)≥a
then a^2+a≥a
then a^2≥0
ii) Suppose a≤0
then 1+a≤1
then a(a+1)≥a since a is negative, by a lemma
then a^2+1≥a
then a^2≥0

2. Apr 16, 2012

### Dick

Seems pretty ok. You aren't stating all of your assumptions and lemmas here, though. And there is a typo in your second argument.

3. Apr 17, 2012

### jaqueh

ok great, yeah i'll be more specific and i do see my typo: a+1 instead of 1+a

4. Apr 17, 2012

### Dick

I actually meant turning a(a+1) into a^2+1, but I'm sure that's not news.

5. Apr 17, 2012

### jaqueh

oh gosh i didn't even realize i had typed that, haha well thanks anyways.

btw you are always incredibly helpful, and for that i am really grateful!

6. Apr 17, 2012

### Curious3141

Why don't you just multiply both sides immediately by a?

i.e.

Case 1: a≥0
a2≥0

Case 2: a≤0
a2≥0

Seems a lot more direct and it's the same idea.

7. Apr 17, 2012

### jaqueh

because you dont know what a*a means yet

8. Apr 17, 2012

### Curious3141

What do you mean, exactly?

9. Apr 17, 2012

### jaqueh

well im taking a beginning real analysis course and basically we're building the basic structure of algebra from scratch from a couple of real number axioms.

10. Apr 17, 2012

### Curious3141

OK. But saying you don't know the implication of a*a means you don't know the implication of a*(a+1) either.

You're already making an implicit assumption when you reverse the sign of the inequality upon multiplying by negative a. Multiplying a≤0 by a assumes no more than multiplying a+1≤1 by a.

11. Apr 17, 2012

### jaqueh

the multiplying done by the negative constant was proved in a lemma but you're right, it wasn't known before. and when i did a(a+1) i guess i knew that it was a*a+a, but i didn't know what a*a was still so i had to describe it using the inequalities