# Math Help - qa is rational if and only if q = 0

1. ## qa is rational if and only if q = 0

This is exactly how the book states it.

Let q ∈ Q and a ∈ R\Q

Does a mean real number? I don't even know what R\Q is..

qa is rational if and only if q=0

2. Originally Posted by glover_m
This is exactly how the book states it.

Let q ∈ Q and a ∈ R\Q

Does a mean real number? I don't even know what R\Q is..

qa is rational if and only if q=0
$\mathbb{R} \backslash \mathbb{Q}$ is the reals with the rationals removed.

So $x \in \mathbb{R} \backslash \mathbb{Q}$ if and only if $x \in \mathbb{R}$ and $x \not\in \mathbb{Q}$.

With that the rest should be easy enough.

CB

3. Originally Posted by CaptainBlack
$\mathbb{R} \backslash \mathbb{Q}$ is the reals with the rationals removed.

So $x \in \mathbb{R} \backslash \mathbb{Q}$ if and only if $x \in \mathbb{R}$ and $x \not\in \mathbb{Q}$.

With that the rest should be easy enough.

CB
yah not really, i figured it was the irrationals, but not sure about proving that heh

the way my teacher does it is he proves both ways.

so in -<<<
Suppose q = 0 = x/y

qa = 0a = 0 =x/y

therefore qa is rational if q=0

but going ->>>>
suppose qa is rational

q=
.
.
.
.
.
=0

that's what i dont understand how to do, i think i understand the first part, but not the second direction.

4. Suppose that q$\neq$0.

Since q is rational and not zero
$q=\frac{x}{y}$
where x$\neq$0.

qa is rational, so

$qa=\frac{z}{w}$
Combining these two we get
$a\frac{x}{y}=\frac{z}{w}\Rightarrow a=\frac{zy}{wx}$
(We can divide by x because it is not zero)
But this means that a is rational, contradiction.
Therefore it must be q=0.

5. Originally Posted by Halmos Rules
Suppose that q$\neq$0.

Since q is rational and not zero
$q=\frac{x}{y}$
where x$\neq$0.

qa is rational, so

$qa=\frac{z}{w}$
Combining these two we get
$a\frac{x}{y}=\frac{z}{w}\Rightarrow a=\frac{zy}{wx}$
(We can divide by x because it is not zero)
But this means that a is rational, contradiction.
Therefore it must be q=0.
You don't really need to go back to the x/y definition of rationals (although you can of course). If q is not 0, then q has a multiplicative inverse and the rationals are closed under multiplication.