$Fix any errors in my proof (beginner)$

Question

$Fix any errors in my proof (beginner)$

Fix any errors in correctness, structure, etc.

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

Proof. By contrapositive, if $a \geq-1$, then it should not be the case that $a < 1/a < b < 1/b$. Consider when a is -1 and b is $\frac{1}{2} $. The statement then would be $-1 < \frac{1}{-1} < \frac{1}{2} < 2 $. In this case, $\neg (a < \frac{1}{a} < b < \frac{1}{b}) $ because $-1 = \frac{1}{-1} $. Therefore, because the contrapositive is true, it must be the case that if $a < 1/a < b < 1/b$, then $a < −1$.

Proofs

Faithalone

361

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.

Answer 1

Answers can be viewed only if

The questioner was satisfied and accepted the answer, or
The answer was disputed, but the judge evaluated it as 100% correct.

View the answer

You can not proof statements just by giving examples. Your proof must be general and work for all possible cases. This is how I would prove it:

Assume
\[(1) a<1/a<b<1/b.\]

Since $a<\frac{1}{a}$, we have
\[a-\frac{1}{a}<0 \Rightarrow \frac{a^2-1}{a}<0.\]
There are two cases that the above inequality can hold:

Case 1. $a^2-1<0$ and $a>0$. Then we must have $0<a<1$. Thus
\[1<\frac{1}{a}<b \Rightarrow b>1.\]
This is a contradiction because if $b>1$, then $b>1/b$ which violates the assumption (1) that $b<1/b$. Thus Case 1 is not possible.

Case 2. $a^2-1>0$ and $a<0$. Then we must have $a<-1$.

Therefore if the assumption (1) holds, then $a<-1$.

Erdos

4.5K

Erdos

0

Please leave a comment if you need any clarifications.
Erdos

0

This took me about 25 minutes to answer. Please consider offering higher bounties for your future posts, otherwise your questions may not get answered.

$Fix any errors in my proof (beginner)$

Answer

Related Questions

Search