qa is rational if and only if q = 0

• February 19th 2009, 10:38 PM
glover_m
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
• February 19th 2009, 11:18 PM
CaptainBlack
Quote:

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
• February 19th 2009, 11:41 PM
glover_m
Quote:

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.
• February 20th 2009, 03:56 AM
Halmos 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.
• February 20th 2009, 05:32 AM
HallsofIvy
Quote:

Originally Posted by Halmos 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.

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.