Prove that $\frac{1}{x}$ does not have an upper bound

Question

I just read at the library how to prove that $\frac{1}{x}$ does not have an upper bound. The idea was to use proof by contradiction. I tried to do the exercise one more time at home, and now I am not sure anymore if my solution is correct. Here is my attempt:

Given the function $f=\frac{1}{x}$ defined on the interval $I=]{0,+\infty}[$

We suppose that this function has an upper bound $M$, which means that $$\forall x \in I \quad \frac{1}{x} \leq M$$

We can consider only the case when $M \gt 0$ because $\forall x \in I$ $\quad \frac{1}{x} \gt0$

Because $M$ is an upper bound, we have $$M+1 \le\frac{1}{x}$$ but that means that $M+1 \le\frac{1}{x}\le M$ which is a contradiction.

Is this correct?

Answer 1

As pointed out in the comments, this is not correct. However, you can take: $$x = \frac{1}{M+1}$$ So that: $$\frac{1}{x}= M+1 > M$$ A contradiction.

Answer 2

Your proof is not correct. $M$ being an upper bound does not imply that $M+1 \leq\frac1x$. Upper bound is not unique. In fact, $M$ being an upper bound implies that $M+1$ is also an upper bound as $M < M+1.$

A correct way would be:

$$\frac1x\leq M\\\frac1M \leq x$$ for all $x>0.$ Thus, $$\frac1M=0$$ which is a contradiction.

Answer 3

The start was OK, we assume $\frac{1}{x}\le M$ for all positive numbers $x$, where $M$ is a positive number. This implies $\frac{1}{M}\le x$ for all positive numbers $x$. But since $\frac{1}{M}$ is positive, we can always find a positive number $x$ with $x< \frac{1}{M}$. So, we get a contradiction.

Answer 4

I think you mean:

Assume $\exists M>0$ such that $$\forall x\in ]0,\infty[: \frac{1}{x}\leq M$$ Let $$x=\frac{1}{M+1}\in]0,\infty[$$ Now $$\frac{1}{x}=M+1\not\leq M\implies\text{contradiction}$$