# Spivak's Calculus, 5(x) - Use (ix) backwards

1. Oct 11, 2008

### Tasaio

Spivak's Calculus, 5(x) -- "Use (ix) backwards..."

1. The problem statement, all variables and given/known data
Prove the following:
(x) If $$a,b\geq0$$ and $$a^{2}<b^{2}$$, then $$a<b$$. (Use (ix), backwards.)

2. Relevant equations

(ix) If $$0 \leq a<b$$, then $$a^{2}<b^{2}$$.

3. The attempt at a solution

Suppose $$a,b\geq0$$ and $$a^{2}<b^{2}$$.

Here's my problem. What does "Use (ix) backwards" mean? I'll assume he means to use the converse of (ix). In that case...

The converse of (ix):

$$\neg(a^{2}<b^{2})\rightarrow\neg(0\leq a<b)$$

Hence $$(a^{2}\geq b^{2})\rightarrow\neg(0\leq a\&\&a<b)$$;

hence $$(a^{2}\geq b^{2})\rightarrow(0>a)\Vert(a\geq b).$$ $$(\star)$$

Since $$a^{2}<b^{2}$$, then $$a^{2} \leq b^{2}$$. So $$b^{2} \geq a^{2}$$.

Then by $$(\star)$$, $$(0>b)\Vert(b \geq a)$$.

Since $$b \geq 0$$, then we know $$0>b$$ cannot be true.

This means that $$b\geq$$ a must be true.

But if $$b=a$$, then $$b^{2}=a^{2}$$; this is a contradiction since we are given that $$a^{2}<b^{2}$$.

Hence b>a must be true.

Hence a<b.

2. Oct 11, 2008

### HallsofIvy

Staff Emeritus
Re: Spivak's Calculus, 5(x) -- "Use (ix) backwards..."

What you have done is perfectly good.