1. Why does adding two even numbers give an even number?

2. Why does adding two odd numbers give an even number?

3. Why does adding an odd and an even number give an odd number?

4. Why aren’t there more even than odd numbers?

I’ll answer each one for you although the last is going to take the most explaining. *grin* So, here goes.

To start with, the answers to these questions are referring only to integers (…, -3, -2, -1, 0, 1, 2, 3, …).

There are several things to keep in mind.

When you add two integers, the answer is always another integer (this is a property of the set of integers called closure).

The associative property – for any integers m, n, p; m + n + p = (m + n) + p = m + (n + p).

The distributive property – for any integers a, b, c; ab + ac = a(b + c).

Now onto the first three questions.

1. Why does adding two even numbers give an even number?

Let’s start with a definition. We are defining an even number as a number that when divided by 2 has no remainder. We can write any even number as 2k where k is any integer. It doesn’t matter if it’s even or odd since multiplying by 2 immediately makes the result (2k) even.

Now add two even numbers. By the definition above, we can choose any two even numbers. Let’s call them 2k and 2m where k and m are integers.

Do the arithmetic: 2k + 2m = 2(k +m) by the distributive property. (k + m) is an integer because adding two integers always results in another integer. Since 2(k + m) is divisible by 2 with no remainder (see our original definition), adding 2k + 2m results in an even number.

2. Why does adding two odd numbers give an even number?

First we’ll define an odd number as a number that when divided by 2 results in a remainder of 1. We know that for any n = 2k, n is even from above. What we want is an n that is odd. By our definition of an odd number, n = 2k + 1 is odd because when we divide n by 2 we end up with a remainder of 1.

Now add two odd numbers. As above, we can choose any two odd numbers. Let’s choose n = 2k + 1 and m = 2r + 1 where k and r are integers.

Next the arithmetic: n + m = (2k + 1) + (2r + 1) = 2k + 2r + 2 (combining like terms).

2k + 2r + 2 = 2(k + r + 1) by the distributive property. Since k, r, and 1 are integers, (k + r + 1) is also an integer. So we now have an even number by our previous definition of an even number.

3. Why does adding an odd and an even number give an odd number?

Choose any odd number m = 2k + 1 (by our previous definition of an odd number) and any even number n = 2p (by our previous definition of an even number) where m, k, n, and p are integers. Now do the addition.

m + n = (2k + 1) + (2p) = 2k + 2p + 1 = (2k + 2p) + 1 (by the associative property) and

(2k + 2p) + 1 = 2(k + p) + 1 (by the distributive property). Since k and p are integers, k + p is an integer, and so we have by definition an odd number.

Now for the last question – why aren’t there more even than odd numbers?

I need to start by talking about infinite sets and how we tell their “size” or cardinality in math speak. The set of natural numbers (counting numbers – 1, 2, 3, …) is an infinite set. It’s a countably infinite set because (at least in theory) you could count all the members of the set. Of course, since there is no “largest” natural number, you’d never actually count to the end. Georg Cantor, when he described infinite sets, said that the set of natural numbers has a cardinality (size) of aleph-0. All countably infinite sets have this cardinality of aleph-0. So all countably infinite sets are the same size. Got that?

In order to show that a set is countably infinite, I have to find a way to match each member of the set to each member of the set of natural numbers. This is a one-to-one correspondence between the two sets. The good thing is that I can rearrange the numbers in the set I’m trying to put into one-to-one correspondence with the set of natural numbers – order, in the every day sense, doesn’t matter in this case.

Let’s start with the even numbers. The set of even numbers (I’m using integers) is (…, -6, -4, -2, 0, 2, 4, 6, …). This is a problem because the even numbers go on forever in the negative and in the positive direction. But I can reorder the set and end up with (0, -2, 2, -4, 4, -6, 6,...).

With this reordering, I can make the match to the set of natural numbers like this (my one-to-one correspondence) by lining up the natural numbers under the even numbers.

0 | -2 | 2 | -4 | 4 | -6 | 6 | and so forth (the even numbers) |

1 | 2 | 3 | 4 | 5 | 6 | 7 | and so forth (the natural numbers) |

Since there is a one-to-one correspondence between the set of even numbers and the set of natural numbers, the set of even numbers is countably infinite and has the same cardinality (size) as the natural numbers – aleph-0.

The odd numbers work the same way.

Reorder the set of odd numbers (…, -5, -3, -1, 0, 1, 3, 5, …) to (0, -1, 1, -3, 3, -5, 5, …).

Now you can make a one-to-one correspondence to the natural numbers in the same way as for the even numbers. With this one-to-one correspondence, the set of odd numbers is countably infinite and so has a cardinality (size) of aleph-0.

Since the set of odd numbers and the set of even numbers both have a cardinality of aleph-0, they are the same size.

Tada!

BTW – the set of rational numbers (your fractions – ½, ¼) is also countably infinite which makes it the same size as the set of integers, the set of even numbers, and the set of odd numbers. The set of real numbers (includes the integers, the rationals, and the irrational numbers) though, is uncountably infinite and has a different cardinality. Proofs are here. I think the proof for the rational numbers is pretty slick.

*Math*

## 4 comments:

You are wonderful. Thanks!!!

Heh. That was actually pretty cool :-)

I knew I could find some math for all of you. *grin*

Thats more then some stuff :P , i stopped reading the blogs when they were ban back here in pakistan , but know i think i should start it again

Post a Comment