Prove that {x∈ℤ : 2|x} ∩ {x∈ℤ : 9|x} ⊂ {x∈ℤ : 6|x}

[lep11]

Homework Statement

Prove that {x∈ℤ : 2|x} ∩ {x∈ℤ : 9|x} ⊂ {x∈ℤ : 6|x}.
It means that the intersection of the two sets is subset of the set containing integers that are divisible by 6.

The Attempt at a Solution

Basically I need to show that if an integer is divisible by 2 and 9, then it's also divisible by 6.
Proof:
let. x=2*9*n=18*n=(6*3*n)∈{x∈ℤ : 6|x} (n∈ℤ)

Q.E.D.

However, I don't think it's that simple.

 

[Samy_A]

Homework Statement

Prove that {x∈ℤ : 2|x} ∩ {x∈ℤ : 9|x} ⊂ {x∈ℤ : 6|x}.
It means that the intersection of the two sets is subset of the set containing integers that are divisible by 6.

The Attempt at a Solution

Basically I need to show that if an integer is divisible by 2 and 9, then it's also divisible by 6.
Proof:
let. x=2*9*n=18*n=(6*3*n)∈{x∈ℤ : 6|x} (n∈ℤ)

Q.E.D.

However, I don't think it's that simple.

You think wrong: it is correct.

 

[lep11]

Ok, thanks

 

[Dick]

You think wrong: it is correct.

Actually, it's correct but needs some justification. Suppose I said x is divisible by 2 and 6. You would be wrong to say that x=2*6*n and conclude that's it's divisible by 12. You need to say something about prime numbers.

 

[lep11]

Actually, it's correct but needs some justification. Suppose I said x is divisible by 2 and 6. You would be wrong to say that x=2*6*n and conclude that's it's divisible by 12. You need to say something about prime numbers.

Let's summarize the proof.

I have already shown that if x is even, then the statement is true.

If x is a prime, additive inverse of a prime or ±1, then {x∈ℤ : 2|x} ∩ {x∈ℤ : 9|x} = the empty set. And the empty set is a subset of every set. In addition, all primes are odd.

Therefore {x∈ℤ : 2|x} ∩ {x∈ℤ : 9|x} ⊂ {x∈ℤ : 6|x} holds for all x∈ℤ.

 

[Samy_A]

Let's summarize the proof.

I have already shown that if x is even, then the statement is true.

If x is a prime, additive inverse of a prime or ±1, then {x∈ℤ : 2|x} ∩ {x∈ℤ : 9|x} = the empty set. And the empty set is a subset of every set. In addition, all primes are odd.

Therefore {x∈ℤ : 2|x} ∩ {x∈ℤ : 9|x} ⊂ {x∈ℤ : 6|x} holds for all x∈ℤ.

I'm a little lost here. The statement "If x is a prime, additive inverse of a prime or ±1, then {x∈ℤ : 2|x} ∩ {x∈ℤ : 9|x} = the empty set" is meaningless, or overuses the symbol "x".

What Dick meant was (I think) that you should explain more in detail why you can write x=2*9*n in your original proof. He gave as an example that this would not work for 2 and 6.

 

[Dick]

I'm a little lost here. The statement "If x is a prime, additive inverse of a prime or ±1, then {x∈ℤ : 2|x} ∩ {x∈ℤ : 9|x} = the empty set" is meaningless, or overuses the symbol "x".

What Dick meant was (I think) that you should explain more in detail why you can write x=2*9*n in your original proof. He gave as an example that this would not work for 2 and 6.

Right, that's what I meant. x being prime isn't an issue, it's something about 2 and 3 being prime.

 

[lep11]

Is it about prime factors? 2*3*3*n? How should I modify the original proof?

 

[Samy_A]

Is it about prime factors? 2*3*3*n? How should I modify the original proof?

Not by much, as both Dick and myself said it is correct.
What you haven't explicitly explained is the following:
Take x ∈{y∈ℤ : 2|y} ∩ {y∈ℤ : 9|y}.
That says that x=2*a and x=9*b, where a and b are integers.
How do you conclude from this that x=2*9*n (n an integer)? We saw that for the pair 2,6 that doesn't work.

 

[lep11]

Isn't that obvious? I still don't get it.

x=2*a and x=9*b
So a=9 and b=2, because x is an integer.

 

[Samy_A]

Isn't that obvious? I still don't get it,

It is rather obvious, but the point was that you didn't state why.

Let's assume I don't get it: I don't see why x=2*a and x=9*b implies x=2*9*n (a, b, n integers). How would you explain it to me?

x=2*a and x=9*b
So a=9 and b=2, because x is an integer.

This is not correct.

 

[lep11]

If x is divisible by 2 and 9, then x must be product of at least 2 and 9.

 

[Samy_A]

If x is divisible by 2 and 9, then x must be product of at least 2 and 9.

Why?
Just start from x=2*a=3*b (a,b integers).
EDIT: sorry, that whas a typo, as above it should have ben x=2*a=3*3*b

 

[lep11]

x=2*a=3*b
a=3b/2

 

[Samy_A]

Let's try this: we have as before x=2*a=9*b.

Do you think that b can be an odd integer? And why?

 

[Dick]

Let's try this: we have as before x=2*a=9*b.

Do you think that b can be an odd integer? And why?

The basic theorem you want to apply here is called "unique prime factorization". Samy A suggests an alternative trick way to do it. But I think you should work with the theorem. Unless you don't know it, in which case the trick is good.

 

[Samy_A]

The basic theorem you want to apply here is called "unique prime factorization". Samy A suggests an alternative trick way to do it. But I think you should work with the theorem. Unless you don't know it, in which case the trick is good.

I agree, and I assume he knows the theorem. But it looks as if he sees the theorem as so obvious that he doesn't actually states it.

 

[lep11]

let x=2*3*3*n=6*3n?