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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!

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    $\begingroup$ Shouldn't the problem statement say: 'prove that ... is irrational'? $\endgroup$ – Toby Mak Feb 23 '20 at 3:57
  • $\begingroup$ you are correct, thanks for pointing that out $\endgroup$ – spuddy Feb 23 '20 at 3:58
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To prove by contradiction, assume the expression is rational with a value of $r$ & manipulate the expression to get

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

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

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

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.

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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 $\frac{1}{\alpha}\not\in\mathbb{Q}$.

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