Proof of $\forall x\in \mathbb{R}^+, \exists q\in \mathbb{R}^+, q<x$.

Question

Prove or disprove the following statement using contradiction: $\mathbb{R}^+=\{x\in \mathbb{R} :x>0\}$, the set of positive real numbers, does not have a minimum element, also called a smallest element.

Here is my answer: Proof: Assume that $\forall q\in \mathbb{R}^+, \exists x\in \mathbb{R}^+, x\geq q$. Let $q$ be a generic element of $\mathbb{R}^+$. Choose $x=\frac{q}{2}$, so $\frac{q}{2}\geq q$, which is a contradiction. Therefore we have proven the original statement, that the set of positive real numbers does not have a minimum element.

Are all of my statements correct? Did I leave out any neccessary rigor? Thanks.

Answer 1

What you've written doesn't make sense but I think you're not far off from grasping the idea of the problem. The statement that the positive real numbers does not have a least element is written like this:

$$\forall x\in \mathbb{R}^+, \exists q\in\mathbb{R}^+, q<x$$

meaning that no matter which positive real number $x$ you choose, I can always find a positive real number $q$ which is strictly smaller than $x$.

We can prove it by contradiction, as you've tried, by assuming the negation holds:

$$\exists x\in\mathbb{R}^+, \forall q\in\mathbb{R}^+, q\geq x$$

Since this statement must hold for all $q\in\mathbb{R}^+$, in particular is holds for $q=\frac{x}{2}$:

$$\frac{x}{2}\geq x$$

But we know that for all positive real numbers we have:

$$\frac{x}{2}<x$$

and thus we have arrived at a contradiction.

Answer 2

The contrary of $$\forall q\in \mathbb R^+, \exists x\in\mathbb R^+: x<q$$ is $$\exists q\in\mathbb R^+: \forall x\in\mathbb R^+, x\geq q.$$ So if you make a proof by contradiction, you suppose that there is a $q\in\mathbb R^+$ s.t. $x\geq q$ for all $x\in\mathbb R^+$. This mean that $q=0$ which is a contradiction with the fact that $q\in\mathbb R^+$.