Let $a,b \in \mathbb{R^+}$; prove that if $ab \geqslant 1$, then $a \geqslant 1$ or $b \geqslant 1$

Question

Is this a valid proof:

We take the contrapositive statement, that if $a < 1$ and $b < 1$, then $ab < 1$, and suppose, to the contrary, that $ab\geqslant 1$. It would follow that \begin{align*} ab && \geqslant && 1\\ a && \geqslant && b^{-1}.\\ \end{align*} By definition, however, $b < 1$ implies $b^{-1} > 1$, and we now have $1 \geqslant a \geqslant b^{-1}> 1$, which compresses to $1>1$, providing a contradiction.

Proof is fine but really really indirect. A bit like making a right turn by doing seven lefts. — fleablood, Sep 28 '17 at 02:19
The contrapositive is a great idea. But if you do it it do a<1 so ab <1b <1. That's all. — fleablood, Sep 28 '17 at 02:29
Proving a contrapositive by contradiction is to assume the initial hypothesis is....true. so you might as well do a direct proof. — fleablood, Sep 28 '17 at 02:46

score 1 · Accepted Answer · 2017-09-28T00:11:55.377

1

Technically yes, this is a valid proof. However it's very clunky and far more complicated than you need.

You proceed by contradicting the contrapositive. You could have completely skipped this step, to say that if $0 < a < 1$, $b \ge a^{-1} > 1$.
There's no need to do contradiction once you've already gotten the contrapositive. If $0 \le a < 1$ and $0 \le b < 1$, then $0 \le ab < 1 \cdot 1 = 1$ and you're already done.

edited Sep 28 '17 at 00:11

answered Sep 27 '17 at 23:21

(+1) Perhaps for beginners, you should insist that it works because $a$ and $b$ are positive.. – Bernard Sep 28 '17 at 00:05
Yes, that's a good point. Thanks. – Sep 28 '17 at 00:06
I don't see how you got to $b=a^{-1}$ – Izzy M. Sep 28 '17 at 00:11
@IzzyM. Because it was a typo. – Sep 28 '17 at 00:12

Let $a,b \in \mathbb{R^+}$; prove that if $ab \geqslant 1$, then $a \geqslant 1$ or $b \geqslant 1$

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