One of the great joys of doing mathematics, and indeed what separates mathematics from all other sciences, is the ability to prove things true beyond a shadow of a doubt. In other sciences, we accept certain laws because they conform to the real world, but those laws can be contra-dicted or modified if new evidence presents itself. But in mathematics, if a statement is proved to be true, then it is true forever. For exam-ple, it was proved by Euclid over two thousand years ago that there are infinitely many prime numbers, and there is nothing that we can say or do that will ever contradict the truth of that statement. Technology comes and goes, but a theorem is forever. As the great mathematician G. H. Hardy said, “A mathematician, like a painter or poet, is a maker of patterns. If his patterns are more permanent than theirs, it is because they are made with ideas.” To me, it often seems like the best route to academic immortality would be to prove a new mathematical theorem.
Not only can mathematics prove things with absolute certainty, it can also prove that some things are impossible. Sometimes people say,
“You can’t prove a negative,” which I suppose means that you can’t prove that there are no purple cows, since one might show up someday.
But in mathematics, you can prove negatives. For example, no matter how hard you try, you will never be able to find two even numbers
that add up to an odd number nor find a prime number that is larger than all other primes. Proofs can be a little scary the first (or second or third) time that you encounter them, and they are definitely an acquired taste. But once you get the hang of them, proofs can be quite fun to read and write. A good proof is like a well-told joke or story—it leaves you feeling very satisfied at the end.
Let me tell you about one of my first experiences of proving some-thing impossible. When I was a kid, I loved games and puzzles. One day, a friend challenged me with a puzzle about a game, and so of course I was interested. He showed me an empty 8-by-8 checkerboard and brought out 32 regular 1-by-2 dominos. He asked, “Can you cover the checkerboard using all 32 dominos?” I said, “Of course, just lay 4 dominos across each row, like this.”
Covering an 8-by-8 checkerboard with dominos
“Very good,” he said. “Now suppose I remove the squares from the lower-right and upper-left corners.” He placed a coin on those two squares so I couldn’t use them. Now can you cover the remaining 62 squares using 31 dominos?”
52
52
Can the checkerboard be covered with dominos when two opposite corners are removed?
“Probably,” I said. But no matter how hard I tried, I was unable to do it. I was starting to think that the task was impossible.
“If you think it’s impossible, can you prove that it’s impossible?” my friend asked. But how could I prove that it was impossible without exhaustively going through the zillions of possible failed attempts? He then suggested, “Look at the colors on the checkerboard.”
The colors? What did that have to do with anything? But then I saw it. Since both of the removed corner squares were light-colored, then the remaining board consisted of 32 dark squares and just 30 light squares. And since every domino covers exactly one light square and one dark square, it would be impossible for 31 dominos to tile such a board. Cool!
Aside
If you liked the last proof, you will enjoy this one as well. The video game Tetris uses seven pieces of different shapes, sometimes given the names I, J, L, O, Z, T, and S.
I J L O
Z T S
Can these 7 pieces be arranged to form a 4-by-7 rectangle?
Each piece occupies exactly four squares, so it is natural to wonder whether it would be possible to arrange them in such a way that they form a 4-by-7 rectangle, where we are allowed to flip or rotate the tiles.
This turns out to be an impossible challenge. How do you prove that it’s impossible? Color the rectangle with fourteen light squares and fourteen dark squares, as shown.
Notice that, with the exception of the T tile, every piece must cover two light squares and two dark squares, no matter where it is located. But the T tile must cover three squares of one color and one square of the other color. Consequently, no matter where the other six tiles are located, they must cover exactly twelve light squares and twelve dark squares. This leaves two light squares and two dark squares to be covered by the T tile, which is impossible.
So if we have a mathematical statement that we think is true, how do we actually go about convincing ourselves? Typically, we start by describing the mathematical objects that we are working with, like the collection of integers
. . .,−2,−1, 0, 1, 2, 3,. . .
consisting of all whole numbers: positive numbers, negative numbers, and zero.
Once we have described our objects, we make assumptions about these objects that we consider to be self-evident, like “The sum or prod-uct of two integers is always an integer.” (In the next chapter, on ge-ometry, we’ll assume statements like “For any two points there is a line that goes through those two points.”) These self-evident statements are called axioms. And from these axioms, and a little logic and algebra, we can often derive other true statements (called theorems) that are some-times far from obvious. In this chapter, you will learn the basic tools of proving mathematical statements.
Let’s start by proving some theorems that are easy to believe. The first time we hear a statement like “The sum of two even numbers is even” or “The product of two odd numbers is odd,” our mind typically checks the statement with a few examples, then concludes that it’s prob-ably true or that it makes sense. You might even think that the statement is so self-evident that we could make it an axiom. But we don’t need to do that, since we can prove this statement using the axioms we already know. To prove something about even and odd numbers, we need to be clear about what odd and even mean.
An even number is a multiple of 2. To express this algebraically, we say n is even if n =2k where k is an integer. Is 0 an even number? Yes, since 0 = 2×0. We are now ready to prove that the sum of two even numbers is even.
Theorem: If m and n are even numbers, then m+n is also an even number.
This is an example of an “if-then” theorem. To prove such a state-ment, we typically assume the “if” part and, through a mixture of logic and algebra, show that the “then” part follows from our assumptions.
Here we assume that m and n are even and want to conclude that m+n is also even.
Proof:Suppose m and n are even numbers. Thus, m=2j and n =2k where j and k are integers. But then
m+n=2j+2k=2(j+k)
and since j+k is an integer, m+n is also a multiple of 2, so m+n must be even.
Notice that the proof relied on the axiom that the sum of two inte-gers (here, j+k) must also be an integer. As you prove more compli-cated statements, you are more likely to rely on previously proved the-orems than the basic axioms. Mathematicians often denote the end of a proof with a designation likeoror Q.E.D on the right margin of the last line of the proof, as seen above. Q.E.D. is the abbreviation for the Latin phrase “quod erat demonstrandum,” meaning “that which was to be demonstrated.” (It’s also the abbreviation for the English phrase
”quite easily done,” if you prefer.) If I think that a proof is particularly elegant, I will end it with a smiley like.
After proving an if-then theorem, mathematicians can’t resist won-dering about the truth of the converse statement, obtained by reversing the “if” part and the “then” part. Here, the converse statement would be “If m+n is an even number, then m and n are even numbers.” It’s easy to see that this statement is false, simply by providing a counterex-ample. For this statement, the counterexample is literally as easy as
1+1=2
since this example shows that it’s possible to have an even sum even when both numbers are not even.
Our next theorem is about odd numbers. An odd number is a num-ber that is not a multiple of 2. Consequently, when you divide an odd number by 2, you always get a remainder of 1. Algebraically, n is odd if n = 2k+1, where k is an integer. This allows us to prove, by simple algebra, that the product of odd numbers is odd.
Theorem: If m and n are odd, then mn is odd.
Proof: Suppose m and n are odd numbers. Then m = 2j+1 and n=2k+1 for some integers j and k. Thus, according to FOIL,
mn= (2j+1)(2k+1) =4jk+2j+2k+1=2(2jk+j+k) +1 and since 2jk+j+k is an integer, this means that the number mn is of the form “twice an integer+1.” Hence mn is odd.
What about the converse statement: “If mn is odd, then m and n are odd”? This statement is also true, and we can prove it using a proof by contradiction. In a proof by contradiction, we show that rejecting the conclusion (here, the conclusion is that m and n are odd) leads to a problem. In particular, if we reject the conclusion, there would also
be a problem with our assumptions, so logically the conclusion must be true.
Theorem:If mn is odd, then m and n are odd.
Proof: Suppose, to the contrary, that m or n (or both) were even. It really doesn’t matter which one is even, so let’s say that m is even and therefore m = 2j for some integer j. Then the product mn= 2jn is also even, contradicting our initial assumption that mn is odd.
When a statement and its converse are both true, mathematicians often call this an “if and only if theorem.” We have just shown the following:
Theorem: m and n are odd if and only if mn is odd.