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

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- Feb 19th 2009, 10:38 PMglover_mqa 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** - Feb 19th 2009, 11:18 PMCaptainBlack
$\displaystyle \mathbb{R} \backslash \mathbb{Q}$ is the reals with the rationals removed.

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

With that the rest should be easy enough.

CB - Feb 19th 2009, 11:41 PMglover_m
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. - Feb 20th 2009, 03:56 AMHalmos Rules
Suppose that qhttp://latex.codecogs.com/gif.latex?\neq0.

Since q is rational and not zero

http://latex.codecogs.com/gif.latex?q=\frac{x}{y}

where xhttp://latex.codecogs.com/gif.latex?\neq0.

qa is rational, so

http://latex.codecogs.com/gif.latex?qa=\frac{z}{w}

Combining these two we get

http://latex.codecogs.com/gif.latex?...=\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. - Feb 20th 2009, 05:32 AMHallsofIvy