$\forall a, b \in \mathbb{R}, \exists M, N \in \mathbb{Z}^*$ such that $Ma=Nb$

Question

Is this true including irrational numbers for $a$ and $b$ (excluding the trivial case of $M = N = 0$)?

Answer 1

There's a trick.

In general it wouldn't be true. If, for example, of $a$ is rational and $b$ is irrational. Then $Ma$ would be rational, and $Nb$ would not be rational, if $N \ne 0$.

ANd if $a,b,N,M$ are not zero then $Ma = Nb \iff \frac ab = \frac NM$ and the RHS must be rational and $a,b$ are under no such obligation.

But the question never said we can't have $M, N$ be $0$.

And if $M=0; N=0$ then $Ma = Nb = 0$ for all $a,b$. So that certainly makes it true.