Proving that the intersection of two sets is a subset of Q?

[Eclair_de_XII]

Homework Statement

"Let ##U={p+r\sqrt{2}:p,r∈ℚ}## and let ##V={a+b\sqrt{7}:a,b∈ℚ}##. Show that ##ℚ⊆U∩V##. Then show that ##U∩V⊆ℚ## and conclude that ##U∩V=ℚ##."

Homework Equations

The Attempt at a Solution

I only knew how to prove the first part. That is:

(1)
"Suppose that ##ℚ⊄U∩V##. This implies that ##∃x∈ℚ## such that ##x∉U## and ##x∉V##. But since ##ℚ⊂U## and ##ℚ⊂V##, it follows that there does not exist an ##x∈ℚ⊂U## such that ##x∉U##. Similarly, there exist no ##x∈ℚ⊂V## such that ##x∉V##."

I'm also ambivalent about saying that ##ℚ⊂U## when I haven't proven yet that ##∃q## such that ##p+q\sqrt{2}∉ℚ##. I'm thinking about just changing it to ##ℚ⊆U##.

(2)
##{p+r\sqrt{2}:p,r∈ℚ}∩{a+b\sqrt{7}:a,b∈ℚ}##

And here is where I got stuck; proving that if ##r≠0##, then ##p+r\sqrt{2}∉V##. Similarly, I was unable to prove that ##a+b\sqrt{7}∉U##. Basically, I'm trying to prove that every element of an intersection whose elements I am unable to properly identify. I'm thinking of just doing it one piece at a time:

"Suppose ##p+r\sqrt{2}∈V##. Then we can say that ##p+r\sqrt{2}=a+b\sqrt{7}##. With algebra, I get: ##\sqrt{2}=\frac{1}{r}(s+b\sqrt{7})## where ##s=a-p##. Squaring both sides gives me ##2=\frac{1}{r}(s^2+2bs\sqrt{7}+7b^2)##. However, contradicts the fact that ##2## is a prime number, since it can only be written as the product of two integers: itself and ##1##."

That's only proving that any element of ##U## with ##r≠0## is an element of ##V##, and I don't even think I'd be done, since I cannot rule out the possibility that ##s+b\sqrt{7}=1## and ##\frac{1}{r}=2##; or the possibility that ##\frac{1}{r}=1## and ##s+b\sqrt{7}=\sqrt{2}##.

I was given a hint that I needed to use the fact that ##\sqrt{14}∉ℚ## for the second part...

 

[Mark44]

Homework Statement

"Let ##U={p+r\sqrt{2}:p,r∈ℚ}## and let ##V={a+b\sqrt{7}:a,b∈ℚ}##. Show that ##ℚ⊆U∩V##. Then show that ##U∩V⊆ℚ## and conclude that ##U∩V=ℚ##."

Homework Equations

The Attempt at a Solution

I only knew how to prove the first part. That is:

(1)
"Suppose that ##ℚ⊄U∩V##. This implies that ##∃x∈ℚ## such that ##x∉U## and ##x∉V##. But since ##ℚ⊂U## and ##ℚ⊂V##, it follows that there does not exist an ##x∈ℚ⊂U## such that ##x∉U##. Similarly, there exist no ##x∈ℚ⊂V## such that ##x∉V##."

I'm also ambivalent about saying that ##ℚ⊂U## when I haven't proven yet that ##∃q## such that ##p+q\sqrt{2}∉ℚ##. I'm thinking about just changing it to ##ℚ⊆U##.

(2)
##{p+r\sqrt{2}:p,r∈ℚ}∩{a+b\sqrt{7}:a,b∈ℚ}##

And here is where I got stuck; proving that if ##r≠0##, then ##p+r\sqrt{2}∉V##. Similarly, I was unable to prove that ##a+b\sqrt{7}∉U##. Basically, I'm trying to prove that every element of an intersection whose elements I am unable to properly identify. I'm thinking of just doing it one piece at a time:

"Suppose ##p+r\sqrt{2}∈V##. Then we can say that ##p+r\sqrt{2}=a+b\sqrt{7}##. With algebra, I get: ##\sqrt{2}=\frac{1}{r}(s+b\sqrt{7})## where ##s=a-p##. Squaring both sides gives me ##2=\frac{1}{r}(s^2+2bs\sqrt{7}+7b^2)##. However, contradicts the fact that ##2## is a prime number, since it can only be written as the product of two integers: itself and ##1##."

That's only proving that any element of ##U## with ##r≠0## is an element of ##V##, and I don't even think I'd be done, since I cannot rule out the possibility that ##s+b\sqrt{7}=1## and ##\frac{1}{r}=2##; or the possibility that ##\frac{1}{r}=1## and ##s+b\sqrt{7}=\sqrt{2}##.

I was given a hint that I needed to use the fact that ##\sqrt{14}∉ℚ## for the second part...

Think about what the elements of ##U \cap V## can be. It's also helpful to think about things from a vector space perspective. If you have a nonzero element q of ##\mathbb{Q}## (a rational number), any other rational number is some (rational) multiple of q. In linear algebra terms, q spans ##\mathbb{Q}##. Set U can be thought of as all linear combinations of elements from ##\mathbb{Q}## and ##\sqrt 2##. With set U, you get all possible rational numbers, plus something else -- all rational multiples of ##\sqrt 2##. Set V is something like this, with all possible rational numbers, plus all rational multiples of ##\sqrt 7##.

So I'm really thinking of U as being spanned by two "vectors" -- 1 and ##\sqrt 2##, and V being spanned by a different set of vectors -- 1 and ##\sqrt 7##. What do the two sets have in common?

 

[vela]

(1)
"Suppose that ##ℚ⊄U∩V##. This implies that ##∃x∈ℚ## such that ##x∉U## and ##x∉V##.

That should be ##x∉U## OR ##x∉V##.

In any case, your proof seems unnecessarily complicated. If ##x \in \mathbb{Q}##, what can you say about ##x + 0\sqrt{2}## and ##x+0\sqrt{7}##?

 

[Eclair_de_XII]

What do the two sets have in common?

They have ##1## as that common element. So how would I express the fact that the sets are in different planes, except at the ##ℚ## axis (so to speak) where they intersect?

If ##x∈ℚ##, what can you say about ##x+0\sqrt{2}## and ##0\sqrt{7}##?

They're both elements of ##U## and ##V##.

 

[vela]

I was given a hint that I needed to use the fact that ##\sqrt{14}∉ℚ## for the second part...

If ##x \in U\cap V##, you have ##x = p+r\sqrt{2} = a+b\sqrt{7}##. Rearrange terms to get ##p-a = b\sqrt{7}-r\sqrt{2}## and then square both sides. See where that takes you.

 

[Eclair_de_XII]

Let's see...

##p^2-2pa+a^2=7b^2-2br\sqrt{14}+2r^2##
##-\frac{p^2-2pa+a^2-7b^2-2r^2}{2br}=\sqrt{14}##

So I get the contradiction that ##\sqrt{14}## is supposed to be irrational? Or is it to early to say that?

 

[Mark44]

Let's see...

##p^2-2pa+a^2=7b^2-2br\sqrt{14}+2r^2##
##-\frac{p^2-2pa+a^2-7b^2-2r^2}{2br}=\sqrt{14}##

So I get the contradiction that ##\sqrt{14}## is supposed to be irrational?

?
##\sqrt{14}## is irrational. The left side, ##p^2 - 2pa + a^2## is rational, as are ##7b^2## and ##2r^2##. Can you have a rational number on one side of an equation, and an irrational one on the other side?

Or is it to early to say that?

 

[Eclair_de_XII]

Can you have a rational number on one side of an equation, and an irrational one on the other side?

No; and that's where I would get the contradiction from. Okay, I think I got my answer. Thanks.