Help with a proof in my discrete math summer class

[CapnCornbread]

Homework Statement

Let A be the set of all integers x such that x is = k2 for some integer k
Let B be the set of all integers x such that the square root of x, SQRT(x), is an integer
Give a formal proof that A = B. Remember you must prove two things: (1) if x is in A, then x is in B, AND (2) if x is in B, then x is in A

Homework Equations

3. Why I am so useless
I am taking an online discrete math class as a prerequisite for my major. The prof is a nice enough person, but the extent of help that he provides on questions that I have asked is "read the book". Which I have...and it didn't help me in this case. If someone could point me in the right direction of how to get this proof underway, I would appreciate it.

 

[LCKurtz]

Homework Statement

Let A be the set of all integers x such that x is = k2 for some integer k
Let B be the set of all integers x such that the square root of x, SQRT(x), is an integer
Give a formal proof that A = B. Remember you must prove two things: (1) if x is in A, then x is in B, AND (2) if x is in B, then x is in A

Since this is your first post, here's a couple of useful hints.

You don't mean what you wrote for A. What you have written would mean A was the even integers. You should write it as k^2 or, better, use the X² key above your edit window. So A = {x: x is an k² for some integer k} which we might more informally call the perfect squares. And B is, informally, the integers that are square roots of other integers.

First you might argue that both sets only contain positive integers. Now suppose x is in A. Can you make an argument that it is in B? Then you have to do the other way.

 

[tiny-tim]

Welcome to PF!

Hi CapnCornbread! Welcome to PF!

Show us your attempt at (1) first. ​

 

[CapnCornbread]

OK, I was thinking I could do something like this:
To prove this example of set equality, we need to undertake two steps: first, show that if x is in A then X is in B, and second, show that if x is in B then X is in A.
First, Assume x ∈A. Then by definition of A, x = k2 for some integer k. Thus by algebra, x = k2 = SQRT(k) * SQRT(k). Therefore x ∈B.
Next, Assume x ∈B. Then by definition of B, SQRT(x) for some integer k. Thus by algebra, x = SQRT(k) = k2. Therefore x ∈A.
We have shown that both A⊆B and B⊆A, therefore A = B.

 

[LCKurtz]

OK, I was thinking I could do something like this:
To prove this example of set equality, we need to undertake two steps: first, show that if x is in A then X is in B, and second, show that if x is in B then X is in A.
First, Assume x ∈A. Then by definition of A, x = k2 for some integer k. Thus by algebra, x = k2 = SQRT(k) * SQRT(k). Therefore x ∈B.

You really need to use the X² icon. You are close, but your equality
k²=SQRT(k) * SQRT(k) isn't true because the left and right sides aren't equal. Think about it a little more.

Next, Assume x ∈B. Then by definition of B, SQRT(x) for some integer k. Thus by algebra, x = SQRT(k) = k2. Therefore x ∈A.
We have shown that both A⊆B and B⊆A, therefore A = B.

Again, sqrt(k) ≠ k². Do you need that?

 

[CapnCornbread]

With revisions:
Assume x ∈A. Then by definition of A, x = k2 for some integer k. Thus by algebra,
x = k2 = k * k. Therefore x ∈B.
Assume x ∈B. Then by definition of B, SQRT(x) for some integer k. Thus by algebra, x = SQRT(k) * SQRT(k) = k. Therefore x ∈A.
We have shown that both A⊆B and B⊆A, therefore A = B.