Prove that $0\le a^2 < b^2$ implies $a < b$

Q) $a$ and $b$ are positive real numbers and $a^2 < b^2$. Prove that $a < b$

My working so far: Assuming always that $a, b > 0$, the contrapositive of $$a^2 < b^2 \implies a < b$$ is

$$a \ge b \implies a^2 \ge b^2\;.$$

Thanks

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HINT: Multiply $a\ge b$ by $a$ to get $a^2\ge ab$ and by $b$ to get $ab\ge b^2$.

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next step: combine, so a^2 > ba > b^2 And a = b implies a^2 = b^2. So together combining gives a≥b and a^2≥b^2. Is this correct? –  DeeDee Jan 9 '13 at 19:53
@DeeDee: You’ve got it. –  Brian M. Scott Jan 9 '13 at 19:58
YES, Thanks for the hint. –  DeeDee Jan 9 '13 at 20:01
@DeeDee: You’re very welcome. –  Brian M. Scott Jan 9 '13 at 20:06

A more direct approach.

If $a,b>0$, then $a+b>0$, and hence $\frac{1}{a+b}>0$.

If $a^2<b^2$ then $$0 < b^2-a^2 = (b-a)(b+a)$$

Multiply both sides by $\frac{1}{a+b}$ to get $$0< b-a$$ so $a<b$.

This shows that you only really need $a+b>0$ and $b^2>a^2$ to show $b>a$.

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Thanks, I understand now –  DeeDee Jan 9 '13 at 20:01
Your contrapositive approach works, as the other answers show, I just liked this answer as a direct approach. (A contrapositive prove is essentially a proof by contradiction, hence sometimes referred to as "indirect.") –  Thomas Andrews Jan 9 '13 at 20:02

$b\le a$ $\implies$ $b^2\le ab\le a^2$

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next step: combine, so a^2 > ba > b^2 And a = b implies a^2 = b^2. So together combining gives a≥b and a^2≥b^2. Is this correct? –  DeeDee Jan 9 '13 at 19:56
@DeeDee edited the question. helped you? –  user52188 Jan 9 '13 at 20:03
Yes, thank you. –  DeeDee Jan 9 '13 at 20:04
@DeeDee $b^2\le a^2$ is already a contradiction. because your assuption is $a^2<b^2$. So you can conclude that you can not have $b\le a$. Therefore $a<b$ –  user52188 Jan 9 '13 at 20:16

I solve it in this way

if $a < b$

by multiplying both side by $a$ .. we get $a^2 < a.b$

,, also by multiplying both side by $b$ .. we get $a.b < b^2$

now we have $a^2 < a.b < b^2$
So $a^2 < b^2$

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For some basic information about writing math at this site see e.g. here, here, here and here. –  Martin Sleziak Nov 13 '13 at 15:31
okey thanks.. i will check it –  Dina Farfosheh Nov 21 '13 at 17:41

This might not be considered a formal proof but it is convincing (because of tis geometrical nature) plus generalises the fact to $(a^n < b^n \Rightarrow a < b \hspace{.5cm} \forall n \in \mathbb{N})$

Think of "$a^n < b^n$" as "the n-volume of a n-cube of side $a$ is smaller than the n-volume of a n-cube of side $b$"

This, as we are talking of n-cubes, certainly means that the n-cube with less volume fits into the other (imagine them sharing one corner) and so the side of the smaller one fits into the side of the bigger one which means $a<b$.

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I know you probably want a trivial proof. But you can see this way: what you want is basically proof that $\sqrt{x}$ is crescent . The derived is positive $\implies$ $\sqrt{x}$ crescent

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