Defining a real number No answers, please, hints only.
I want to express every nonzero $x\in \mathbb R$ as a product of two numbers that are not rational.
My attempt is $x=(a_1+b_1i)(a_2+b_2i)$ where $a_1, b_2\in \mathbb R$
Am I correct or do I need a hint?
 A: Hints:

*

*Why is addressing the case with positive $x\ne1$ the hard part?

*For such $x$, consider rational and irrational $x$ separately, and think about how $n$th roots would help.

A: Hint. Play with examples.

*

*$2=\sqrt{2}\cdot \sqrt{2}$

*$-4=\frac{-4}{\sqrt{3}}\cdot \sqrt{3}$.

*$\frac{7}{5}=\sqrt{\frac{7}{5}}\cdot\sqrt{\frac{7}{5}}$

*$\sqrt{2}=\frac{\sqrt{2}}{2}\cdot 2\sqrt{2}$

*etc.

Now generalize...
A: How many ways are there to express $x=ab$ as product of two reals? How many of these ways have rational $a$? How many have rational $b$?

More constructively: Can you achieve $a=\sqrt 2$? Can you achieve $a=\sqrt 3$? Could it happen that $b$ is rational in both cases?
A: *

*If $\frac{x}{\sqrt 2}$ is irrational we are done;

*If $\frac{x}{\sqrt 2}$ is rational. Let $x = \sqrt 2 \cdot \frac mn$ where $\gcd(m,n)=1$, and $$y = \frac{\sqrt 2}{\sqrt{4n^2+4n-1}}, z = \frac{m}{n}\cdot \sqrt{4n^2+4n-1} $$ Then $x=yz$ where $y$, $z$ are both irrational. In fact $4n^2+4n-1$ can be replaced with any non-square integer coprime to both $2$ and $n$, for example any prime number greater than both $n$ and $2$.

A: A bit of cardinality might help.
For a given $x$ consider all then $w \in \mathbb R$ so that $w \ne 0$.
Only a countable number of $w$ will give us $\frac xw$ is rational, and an uncountable number of $w$ will give us $\frac xw$ is irrational.  And only a countable number of $w$ are rational and an uncountable number of $w$.  So there are an uncountable number of $w$ that are irrational and and have $\frac xw$ are irrational.
So just pick one of those.  Then $w$ is irrational and $\frac xw$ is irrational and $x = \frac xw\cdot w$.
So we know it can always be done.  But how to do it?
Well, if $x$ is rational, then we can pick $w$ be any irrational, say $w=\pi$.  Then $\pi$ is irrational, and $\frac x{\pi}$ is irrational, and $\pi\cdot \frac x\pi = x$.
ANd if $x$ is irrational.... well either $\frac x\pi$ is rational or irrational.  If $\frac x\pi = q\in \mathbb Q$ then $x = \pi*q$.  So we need to pick an irrational $w$ so that $\frac xw = \frac {\pi *q}{w}$ is irrational.
Well there are an uncountable number of irrational $w$ so that $\frac \pi w$ is irrational.  Can we come up with one?  Well $\sqrt \pi$ is irrational and $\frac {\pi}{\sqrt \pi} = \sqrt \pi$ is irrational.
So
If $\frac x{\pi}\not \in \mathbb Q$ let $w = \frac x{\pi}$ and $\pi \cdot \frac x{\pi} = x$.
If $\frac x{\pi} \in \mathbb Q$ then $\frac x{\sqrt{\pi}}\not \in \mathbb Q$.  Then $\sqrt \pi\cdot \frac x{\sqrt \pi} = x$.
That'll do it.
