Odd and Even Multiplication

Date: 02/17/99 at 15:32:46
From: Ben Bloom
Subject: Odd and even Multiplication

We think you always get an even answer when you multiply two even 
numbers or an even number and an odd number. We also think that you 
always get an odd answer when you multiply two odd numbers. Do you 
know any examples when this is not true? 

Mr. Blackwell's 4th grade class

Date: 02/17/99 at 16:33:47
From: Doctor Rob
Subject: Re: Odd and even Multiplication

You are correct! All three statements about multiplication are true. 
There cannot be any examples where your statements are not true. You 
can see that this is true as follows.

Notice that the last digit of a product depends only on the last digit 
of the two numbers multiplied together. This is because of the way you 
multiply numbers with more than one digit together. That means that 
your statements are true if they are true for one-digit numbers. Now 
look in your multiplication table up to 9-times-9, and check that they 
are true there, for all one-digit numbers.

- Doctor Rob, The Math Forum
 http://mathforum.org/dr.math/   

Date: 02/17/99 at 17:11:57
From: Doctor Peterson
Subject: Re: Odd and even Multiplication

No, there are no counterexamples for your conjecture. The fun part of 
math is to go one step beyond a conjecture and see why it has to be 
true, not just for every example you can find, but for every case. Let 
us see if we can prove that this is always true, without using any 
complicated ideas that you haven't learned yet.

I think you can see it with a picture. First, remember that a number 
is even if it can be arranged like this (using some sort of counters):

o o o o o o o  14 is even
o o o o o o o

rather than like this:

o o o o o o o  13 is odd
o o o o o o

That is, if the counters can be paired off with nothing left over, the 
number is even.

Now remember that you can multiply two numbers by making a rectangle, 
with one number for the height and the other number for the width:

o o o o o o o  3 x 7 = 21
o o o o o o o
o o o o o o o

Now, suppose the width is even, like this:

o o o o o o   5 x 6 = 30
o o o o o o
o o o o o o
o o o o o o
o o o o o o

Then you can always pair off all the counters in the rectangle by 
pairing off columns:

o o|o o|o o
o o|o o|o o
o o|o o|o o
o o|o o|o o
o o|o o|o o

It doesn't matter whether the height is odd or even. So an even number 
times an odd number is always even. But suppose you multiply an odd 
number by an odd number. Let's try pairing it off. First we can pair 
off the columns as far as they go:

  o o|o o|o o|o  5 x 7 = 35
  o o|o o|o o|o
  o o|o o|o o|o
  o o|o o|o o|o
  o o|o o|o o|o

Since the width is odd, we know we have a single column left over. We 
can pair those counters off as far as they go:

o o|o o|o o|o o
o o|o o|o o|
o o|o o|o o|o o
o o|o o|o o|
o o|o o|o o|o

Since the height is odd, we know we will have one left over - so the 
product of an odd number times an odd number is always odd.

Do you see that the particular example I used does not matter? At 
every step I could say, "we know this will always be true" because my 
thinking did not depend on anything that might be different from one 
example to another. That's how a mathematician thinks: we can prove 
that something is always true by making sure that everything we say 
only depends on what we are assuming, in this case that both numbers 
are odd.

In case you are interested, here is how I can prove the same thing 
without using counters:

Any even number can be written as 2xN for some number N. Any odd 
number can be written as 2xN + 1 for some number N.

The product of an even number 2xN and any other number M is

(2xN)xM = 2x(NxM)

(this is called the associative property), so it is 2 times something: 
it is even.

The product of an odd number 2xN + 1 and another odd number 2xM + 1 is

(2xN + 1)x(2xM + 1)
 = (2xN)x(2xM + 1) + 1x(2xM + 1)
 = 2x(2xNxM + N) + 2xM + 1
 = 2x(2xNxM + N + M) + 1

(which you may or may not be able to follow!), so it is one more than 
a multiple of 2, and must be odd.

By using symbols instead of models, I can more easily convince myself 
it is always true, but it is really exactly the same reasoning either 
way.

Now that I have proved your conjecture, I don't have to try to find a 
counterexample. That saves a lot of work, proving MY conjecture: 
mathematicians are lazy!

- Doctor Peterson, The Math Forum
 http://mathforum.org/dr.math/