# Help with an evidence : Given $a\ne 0$ and also $b\ne 0$ and also $a \lt \frac{1}{a} \lt b \lt \frac{1}{b}$. Confirm $a\lt -1$

This inquiry is from guide How to Prove It and also I'm having actually problem getting going with it. Guide gives the tip "first confirm $a \lt 0$". Nonetheless, I can not identify just how to get that much with the given givens.

The inquiry is:

Given $a\ne 0$ and also $b\ne 0$ and also $a \lt \frac{1}{a} \lt b \lt \frac{1}{b}$. Confirm $a \lt -1$.

Hint: What does a < 1/ a claim concerning a? What does b < 1/ b claim concerning b?

Suppose that $a$ declared. After that, increasing the string of ineqs. (does not transform their instructions) by $a$ offers $a^2 <1 < ba < a/b$. From below we see that $b>1$ (why?), yet after that, from the last ineq., $b^2 a < a$ which is difficult (why?). Consequently $a<0$. Currently increase once more by $a$ and also end.

**HINT ** $\ \ \ $ Multiplying $\rm\ a < 1/a\ \:$ by $\rm\ a^2\ $ returns $\rm\:\ a^3 < a\ $. By proportion $\rm\ b^3 < b\ $. So both $\rm\: a\:$ and also $\rm \:b\:$ have to hinge on the periods where the chart of $\ x^3 - x < 0\:,\ $ i.e. either in $(-\infty,-1)$ or in $(0,1)\:$. Yet $\rm\ a \in (0,1)\ \Rightarrow\ 1/a > 1\ $ opposite $\rm\ 1/a < b < 1\ $.

$\ $ If $\rm\ a > 0\ $ after that $\rm\ a < b\ $ times $\rm\ 1/a < 1/b\ \Rightarrow\ 1 < 1$

So $\rm\ a < 0\ \:$ and also $\rm\:\ a < 1/a\ \:\Rightarrow\:\ a^2 > 1 \ \:\Rightarrow\:\ a < -1$

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