# How do I prove by contradiction?

I'm stuck on this question and I don't know how where to start.

Let $$a$$ and $$b$$ be rational numbers with $$a$$ is not equal to $$b$$

Prove that $$a+\frac{b-a}{\sqrt2}$$ is irrational. (You may assume $$\sqrt2$$ is irrational)

Hence, prove that there is an irrational number between any two rational numbers.

Any help is greatly appreciated!

• Shouldn't the problem statement say: 'prove that ... is irrational'? – Toby Mak Feb 23 '20 at 3:57
• you are correct, thanks for pointing that out – spuddy Feb 23 '20 at 3:58

To prove by contradiction, assume the expression is rational with a value of $$r$$ & manipulate the expression to get

\begin{aligned} r & = a + \frac{b-a}{\sqrt{2}} \\ r - a & = \frac{b-a}{\sqrt{2}} \\ \sqrt{2}(r - a) & = b-a \\ \sqrt{2} & = \frac{b-a}{r - a} \end{aligned}\tag{1}\label{eq1A}

Note since $$b \neq a$$, then $$r - a \neq 0$$ so the division by it is allowed in the last line.

Since $$b$$ and $$a$$ rational, then so is $$b - a$$. Similarly, by the assumption $$r$$ is rational, then so is $$r - a$$. As the ratio of $$2$$ rational values is rational, this means that $$\sqrt{2}$$ is rational. However, as it's actually irrational, this means our original assumption is incorrect, so $$r$$ must be irrational instead.

As for the second part, assume $$b \gt a$$. Then $$r \gt a$$. Also, note that

\begin{aligned} b - r & = b - a - \frac{b-a}{\sqrt{2}} \\ & = (b - a)(1 - \frac{1}{\sqrt{2}}) \\ & = (b - a)(\frac{\sqrt{2} - 1}{\sqrt{2}}) \end{aligned}\tag{2}\label{eq2A}

Since $$\sqrt{2} \gt 1$$, you have that $$b - r \gt 0 \implies r \lt b$$. Put together, you get

$$a \lt r \lt b \tag{3}\label{eq3A}$$

You can get a similar result if you assume $$a \lt b$$ instead. Thus, this shows there's always an irrational between any $$2$$ rational values.

Suppose $$a+\frac{b-a}{\sqrt{2}}\in\mathbb{Q}$$. Then we can write $$a+\frac{b-a}{\sqrt{2}}=r$$, where $$r$$ is a rational number. Since $$a$$ is also rational, we get $$r-a$$ is rational, that is,

$$\frac{b-a}{\sqrt{2}}=r-a\in\mathbb{Q}.\quad(1)$$

We know, by hypotesis, that $$a,b\in\mathbb{Q}$$ and $$b\neq a$$ so this relation implies $$0\neq b-a\in\mathbb{Q}$$. Being $$r-a$$ and $$b-a$$ rational numbers, $$\frac{r-a}{b-a}$$ is also a rational number. Thus equation (1) gives $$\frac{1}{\sqrt{2}}=\frac{r-a}{b-a}\in\mathbb{Q}$$ and this is a contadiction since $$\alpha\not\in\mathbb{Q}$$ iff $$\frac{1}{\alpha}\not\in\mathbb{Q}$$.