Proof with even numbers and divison

In summary: And we want to prove:4|(a.b) ... (2)So, we can use (1) to prove (2). (We'll do this by contradiction)Assume, by contradiction, that 4 doesn't divide a.b. It means, a.b is not a multiple of 4. Now, since a.b is not a multiple of 4, 4 must divide only one of a, or b (This is because if 4 divides both a, and b, then 4 will divide a.b). Let's suppose, 4 divides a. Now, we know that, a = 2n, for some integer n. So, 4|a means, 4|
  • #1
gmmstr827
86
1

Homework Statement



Prove that the product of two even integers is divisible by four.

Homework Equations


The Attempt at a Solution



If a·b where (a^b)ЄZ+, then 4|(a·b)
If 2|a ^ 2|b, then (a^b)≥2 v (a·b)=0

If 2|(2·n) then 4|[2·(2·n)] for nЄZ

∴ 4|(a·b) where (a^b)ЄZ+
[X]Is my proof correct? Is my notation for a positive integer correct (Z+), or should I not use the + and instead state that a,b≥0 a,bЄK? I'm not sure if that really proves it or not, but it's the best I have at the moment.

I was given much shorter proofs after I constructed this, and am aware it can be much shorter, but I mainly want to know if this is CORRECT or not.

Thank you!
 
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  • #2
gmmstr827 said:

Homework Statement



Prove that the product of two even integers is divisible by four.

Homework Equations


The Attempt at a Solution



If a·b where (a^b)ЄZ+, then 4|(a·b)

What do you mean by "If a.b"? And your notation is invalid. You cannot say that a, and b are positive integers by writing (a^b)ЄZ+.

The symbol ^ is used to connect 2 statements. a, and b are 2 integers, not statements, so, you cannot write like that.

If 2|a ^ 2|b, then (a^b)≥2 v (a·b)=0

If 2|(2·n) then 4|[2·(2·n)] for nЄZ

n seems to be jumping out of nowhere. When writing a formal proof, everything must be defined. What is n doing there? What's the relation between n, and the existing variable(s)?

∴ 4|(a·b) where (a^b)ЄZ+
[X]

And what's the connection between n, and a, b?
 
  • #3
I was told ^ simply means "and" so I thought it would be appropriate.
a·b is a common way to say "a multiplied by b." Even calculators use the notation of a centered dot to represent multiplication.
I first showed that a and b are both divisible by 2 since they are even.
I used n as an integer separate from the positive number definition to show that any even number is divisible by 2 and therefore the multiplication of two even numbers is divisible by 4.
 
  • #4
gmmstr827 said:
I was told ^ simply means "and" so I thought it would be appropriate.
a·b is a common way to say "a multiplied by b." Even calculators use the notation of a centered dot to represent multiplication.

Well, yes, I know it's multiplication sign, but why are you putting 'If' in front of a.b? It doesn't make much sense.

I first showed that a and b are both divisible by 2 since they are even.

No, you don't need to show that, it's how even numbers are defined to be.

Even integers are integers that are divisible by 2. So if a, and b are even integers, then they are divisible by 2. You don't need to prove it.

I used n as an integer separate from the positive number definition to show that any even number is divisible by 2 and therefore the multiplication of two even numbers is divisible by 4.

You must introduce the appearance of n, like this:

[tex]a \mbox{ is an even integer } \Rightarrow \exists n \in \mathbb{Z} : a = 2n[/tex]
[tex]b \mbox{ is an even integer } \Rightarrow \exists m \in \mathbb{Z} : b = 2m[/tex]

":" stands for 'such that'
[tex]\exists[/tex] means 'there exists'.

One important thing you should note is that, in a formal proof, you should always defined/introduce every single variable. How they are defined, where they come from, blah blah blah. You cannot just throw them out of nowhere.
 
  • #5
The first line I used to state mathematically what the problem said, which is why I said if, because it has to exist. However, now I believe ∃ would probably be a better way to start it. So, how does this look?

The following is simply how I restate the problem, saying it exists, that they are integers, and that they are even. I believe the ^ is appropriate here since those are statements?
∃ 4|(a·b) and 2|a ^ 2|b : a,bЄZ

Now I introduce n and m. Since I defined a and b as even in the last step, I believe it would be unnecessary here?
a → ∃n Є Z : a = 2·n
b → ∃m Є Z : b = 2·m

I show that 4|[(2·n)(2·m)] relates to 4|(a·b):
4|[(2·n)(2·m)]
4|(4·n·m) → 4|(a·b)

Therefore;
∴ 4|(a·b)
[X]

Is anything wrong with this proof?
Is the notation ЄZ+ correct for positive integers? (I at first read the question wrong and thought it said POSITIVE even integers, which is why I tried incorporating that, but I'm curious if it's correct for future use).
Apologies if it seems I'm making obvious mistakes, I'm new to writing proofs and this is the hardest one I've yet to write since the class pretty much just started. I'm not yet familiar with the notation of things.
 
  • #6
gmmstr827 said:
The first line I used to state mathematically what the problem said, which is why I said if, because it has to exist. However, now I believe ∃ would probably be a better way to start it. So, how does this look?

The following is simply how I restate the problem, saying it exists, that they are integers, and that they are even. I believe the ^ is appropriate here since those are statements?
∃ 4|(a·b) and 2|a ^ 2|b : a,bЄZ

No, that's not how you would use 'exists'. 'exists' ([tex]\exists[/tex]), and 'for all' ([tex]\forall[/tex]) only goes with variables. 4|(a.b) is a statement (4 divides a.b).

Here's some ways you can use 'exists', and 'for all' in making statements.

  • [tex]a \vdots 3 \Rightarrow \exists k \in \mathbb{Z} : a = 3k[/tex]
  • [tex]x \ge 0, \forall x \in \mathbb{N}[/tex]
  • [tex]\exists x \in \mathbb{Z} : x + 3 = 0[/tex]

-------------------------------------

If you want to write down what your problem asks Mathematically, I'd write like this:
[tex]\left. \begin{array}{l} a \mbox{ is even} \\ b \mbox{ is even} \end{array} \right\} \Rightarrow 4|(a.b)[/tex]

Now I introduce n and m. Since I defined a and b as even in the last step, I believe it would be unnecessary here?
a → ∃n Є Z : a = 2·n
b → ∃m Є Z : b = 2·m

This looks not bad.

I show that 4|[(2·n)(2·m)] relates to 4|(a·b):
4|[(2·n)(2·m)]
4|(4·n·m) → 4|(a·b)

You can write it a little bit reverse. We always go from what we know to be true, to what we need to prove. Like this:

We have:
[tex]4|(4.m.n)[/tex] (this is because there's a factor 4 in 4.m.n, hence 4 divides 4.m.n)
[tex]\Rightarrow 4|((2.m).(2.n))[/tex]
[tex]\Rightarrow 4|(a.b)[/tex]

Is the notation ЄZ+ correct for positive integers? (I at first read the question wrong and thought it said POSITIVE even integers, which is why I tried incorporating that, but I'm curious if it's correct for future use).

Yup, [tex]\mathbb{Z} ^ {+}[/tex] stands for the set of positive integers.

Apologies if it seems I'm making obvious mistakes, I'm new to writing proofs and this is the hardest one I've yet to write since the class pretty much just started. I'm not yet familiar with the notation of things.

Everything is hard in the beginning, but I assure you it'll get easier as you practice more and more. Don't worry. Just try our best, and you'll get what you want. :)
 

1. What is the definition of an even number?

An even number is a whole number that can be divided evenly by 2, with no remainder. This means that when an even number is divided by 2, the result will always be another whole number.

2. How do we prove that a number is even?

To prove that a number is even, we can use the division method. If the number can be divided evenly by 2, then it is an even number. Another way to prove it is by using the definition of even numbers, which is that an even number is always one more than an odd number.

3. Can we use any number to divide an even number?

Yes, any number can be used to divide an even number, as long as there is no remainder. This is because the definition of an even number is that it is divisible by 2.

4. How do we use proof by division to show that a number is even?

To use proof by division, we divide the given number by 2. If there is no remainder, then the number is even. Another way to use proof by division is to show that the given number is one more than an odd number, since even numbers are always one more than an odd number.

5. Can we use the same method to prove that a number is odd?

No, we cannot use the same method to prove that a number is odd. To prove that a number is odd, we use the division method and show that the number divided by 2 results in a remainder of 1. We can also use the definition of odd numbers, which is that they are always one less than an even number.

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