# Thread: Proof by Construction.

1. ## Proof by Construction.

I need to prove using the method of proof by construction that: Between every two distinct rational numbers lies an irrational number.

I started my proof assuming that $x$ and $y$ are 2 rational numbers such that $x.

Since $x$ and $y$ are rational

Therefore $\exists_a, _b, _c, _d\in Z$ with $b$ and $d$ not = 0 such that $x = \frac{a}{b}$ and $y = \frac{c}{d}$

I need to construct an element q that can be between x and y and that has the properties of an irrational number. I thought about using the geometric mean of the numbers x and y: $\sqrt{xy}$ but the geometric mean can be a rational number or an irrational number. I don't know if I have to prove this element q to be irrational all the time.

Any assistance would be greatly appreciated, thank you.

2. Suppose that $\left\{ {a,b} \right\} \subseteq \mathbb{Q}\,\& \,a < b$ it follows that $\sqrt 2 a < \sqrt 2 b$.
From that it follows $\left( {\exists r \in \mathbb{Q}} \right)\left[ {\sqrt 2 a < r < \sqrt 2 b} \right]$.
Can you finish?

3. When you say that $\left\{ {a,b} \right\} \subseteq \mathbb{Q}\,\& \,a < b$ you are referring to two different numbers then when i say that $x = \frac{a}{b}$ right?

Originally Posted by Plato
Suppose that $\left\{ {a,b} \right\} \subseteq \mathbb{Q}\,\& \,a < b$ it follows that $\sqrt 2 a < \sqrt 2 b$.
From that it follows $\left( {\exists r \in \mathbb{Q}} \right)\left[ {\sqrt 2 a < r < \sqrt 2 b} \right]$.
Can you finish?

4. And if that was the case, would you just have to divide each number by $\sqrt{2}$ so that you would have $a<\frac{r}{\sqrt{2}} and since $\frac{r}{\sqrt{2}}$ is an irrational number the construction is complete?

5. Originally Posted by xwanderingpoetx
since $\frac{r}{\sqrt{2}}$ is an irrational number the construction is complete?
Well of course!