Proving Rational Intersection of Sets with Irrational Elements

[KeynesianDude]

Homework Statement

Let S={p+q$$\sqrt{2}$$ : p,q $$\in$$ Q} and T={p+q$$\sqrt{3}$$ : p,q $$\in$$ Q}. Prove that S$$\cap$$T = Q.

Homework Equations

See above.

The Attempt at a Solution

I was thinking possible using

S$$\cap$$T=Q
S + T - S$$\cup$$T = Q

But I have no idea how to combine them? I don't believe it's necessary to first prove S and T are rational individually.

 

[lanedance]

hmmm... not sure but is there some way to show that

$$p + q\sqrt{2} = s + t\sqrt{3}$$
iff q=t=0
(maybe consider the case p = s = 0 first)

then the intersection reduces to only the rationals?

 

[KeynesianDude]

Thanks for your response. I was thinking the same thing.

Would it be a bad idea to transfer this to one of the math discussion threads? It's high-level undergrad and low-level grad type proof writing class.

 

[lanedance]

why not try it first... (more volume here & they're not meant for homework)
say you have
$$p + q\sqrt{2} \in Q}$$
$$s + t\sqrt{3} \in T}$$

the intersection will be given by elements that satisfy
$$p + q\sqrt{2} = s + t\sqrt{3}$$

re-arranging
$$p-s = (t\sqrt{3} -q\sqrt{2})$$

the left is cleary rational, can the righthand side ever be rational? (haven't tried to prove it or find conditions, but probably a good place to start)

 

[Dick]

why not try it first... (more voluem here & they're not meant for homework)
say you have
$$p + q\sqrt{2} \in Q}$$
$$s + t\sqrt{3} \in T}$$

the intersection will be given by elements that satisfy
$$p + q\sqrt{2} = s + t\sqrt{3}$$

re-arranging
$$p-s = (t\sqrt{3} -q\sqrt{2})$$

the left is cleary rational, can the righthand side ever be rational? (haven't tried to prove it or find conditions, but probably a good place to start)

You should probably try and solve it before you transfer it. lanedance has good suggestions. Square both sides if t and q are both nonzero. Is sqrt(6) rational? Or is sqrt(2) or sqrt(3) if one of t or q happen to be zero?

 

[KeynesianDude]

You should probably try and solve it before you transfer it. lanedance has good suggestions. Square both sides if t and q are both nonzero. Is sqrt(6) rational? Or is sqrt(2) or sqrt(3) if one of t or q happen to be zero?

[$$\sqrt{6}$$ is definitely not rational. I squared both sides and am kind of clueless. It seems like no matter what I do it will always come off as irrational.

It seems like I need to get t or q to be zero to this to work, huh?

 

[Dick]

You want to prove t AND q are both zero. You've got three cases to worry about. t=0, q not zero, q=0, t not zero and q and t both not zero. Yes, you need t=q=0 to make it work. Can you show this?

 

[KeynesianDude]

You want to prove t AND q are both zero. You've got three cases to worry about. t=0, q not zero, q=0, t not zero and q and t both not zero. Yes, you need t=q=0 to make it work. Can you show this?

Would this be acceptable?

Suppose 3t²-2$$\sqrt{6}$$tq+3q is rational, then q:=0. Hence,

(p-s)²=3t²

Both sides are clearly rational. Therefore, S$$\cap$$T $$\in$$ Q.

 

[lanedance]

helps if you show what you are doing, but i think that's the plan

Remember this is the equation that must be sastified by any points in the intersection of S & T.

so note if u is rational, then so is u^2. So as you're hinting... what limits can you put on t & q, to be in the intersection? will need to consider all the cases as Dick mentioned

 

[lanedance]

Would this be acceptable?

Suppose 3t²-2$$\sqrt{6}$$tq+3q is rational, then q:=0. Hence,

(p-s)²=3t²

Both sides are clearly rational. Therefore, S$$\cap$$T $$\in$$ Q.

no - consider the case separately when t,q both non zero, then look at the square
if one of t,q = 0, there is no need to look at the square

note that sqrt(2) is not rational but (sqrt(2))^2 is not,
i think this is because you can show that u^2 irrational implies u is irrational, but u irrational does not imply u^2 is irrational