Prove, by contradiction, that if a and b are nonzero real numbers, and $a<\frac{1}{a}<b<\frac{1}{b}$ then $a<−1$.

Question

Prove, by contradiction, that if a and b are nonzero real numbers, and $a<1/a<b<1/b$ then $a<−1$.

I understand that the first step is to assume that $a<1/a<b<1/b$ is true. Therefore, the hypotheses makes up:

$a ≠ 0$

$b ≠ 0$

$a<1/a<b<1/b$ is true

From here, to prove by contradiction, we find the negated version of $ a < -1$ which is $ a => -1 $.

Now we must find a contradiction within these statements, but I am struggling to pinpoint that contradiction.

FYI Answer: I have accepted the completed answer for those who are looking for a quick answer but do check the other one for in-depth analysis as well.

Answer 1

suppose $-1\le a$, i.e $a\in [-1,\infty)$.

Note that $a$ cannot be larger than $1$ or be in $(-1,0)$, otherwise, $\frac{1}{a}<a$. And $a$ cannot be $-1$ or $1$. otherwise, $\frac{1}{a}=a$. Therefore, $a\in(0,1)$ and $1<\frac{1}{a}$

But, there is $b$ such that $1<\frac{1}{a}<b$, thereofre, $\frac{1}{b}<1<b$, which contradicts to the assumption that $b<\frac{1}{b}$.

Answer 2

The trick is to note that $1$ is always between $|c|$ and $\frac 1{|c|}$ and So if $c \ne \pm 1$ then either you have have $0 < c < 1 < \frac 1c$ or $0< \frac 1c < \pm < c$ or $c < -1 < \frac 1c < 0$ or $\frac 1c < -1 < c < 0$.

So if $a < \frac 1a < b < \frac 1b$ then $a < \pm 1 < \frac 1a < b < \pm 1 < \frac 1b$.

Now there are only two possible $\pm 1$, $-1, 1$ and $-1< 1$ so si you have $\pm 1 < \pm 1$ the only true statement that makes sense is $-1 < 1$ so $a < -1 < \frac 1a < b < 1 < \frac 1b$.

and ... okay that was a direct proof and you were asked for a proof by contradiction:

......

Note that If $a < 0$ then $a < \frac 1a \implies a^2 > 1$ and $|a| >1$ so if $a<0$ then $a<-1$.

As we are assuming both $a \ge -1$ and $a < \frac 1a$ so that means $a \ge 0$ and as $a\ne 0$ we have $a > 0$. So $a< \frac 1a \implies a^2 < 1$ and $|a| < 1$ so $0 < a < 1$. Which means $\frac 1a > 1$

So we have $0 < a < 1 < \frac 1a$.

Can you finish?

If we assume $0 < a < 1 < \frac 1a < b < \frac 1b$ then $b > 1>0$ and .....