Start from what you are given
A direct proof of "if P then Q" assumes P and reaches Q by steps that are each forced by what came before. The first step is to write P in a form you can work with, usually by unfolding a definition.
Take the claim "the sum of two odd numbers is even". As a conditional it says: if a and b are odd, then a + b is even. An odd number is one more than an even number, so write a = 2m + 1 and b = 2k + 1, where m and k are whole numbers.
The two odd numbers need two different letters, m and k. Writing both as 2k + 1 would make a and b the same number, and the proof would then cover only an odd number added to itself.
Forced steps
Now add the two numbers, and justify each line by algebra alone. Start with a + b = (2m + 1) + (2k + 1).
Collect the terms: a + b = 2m + 2k + 2. Take out the common factor 2: a + b = 2(m + k + 1).
Nothing on the way assumed that a + b is even. Each line follows from the one above it, so whatever was true of a and b at the start is still true at the end.
Reaching the conclusion
The number m + k + 1 is a whole number, because m and k are. So a + b is 2 times a whole number, and that is the definition of even. The conclusion is reached, and the proof ends.
For example, 7 = 2 × 3 + 1 and 9 = 2 × 4 + 1, so m = 3 and k = 4. Then 7 + 9 = 16, and 2(3 + 4 + 1) = 2 × 8 = 16.
Three pairs of odd numbers, with a = 2m + 1 and b = 2k + 1. In every row a + b is twice m + k + 1, as the proof says. The rows agree with the proof, but only the algebra covers every pair.
The whole method
Start: assume the hypothesis and write it in symbols. Middle: take steps that each follow from what came before, with nothing else assumed. End: show that the last line is the conclusion, usually by matching it to a definition.
The same three moves prove that the sum of two even numbers is even. Write the two numbers as 2a and 2b, where a and b are whole numbers. Their sum is 2a + 2b = 2(a + b), which is 2 times a whole number, so it is even.
They prove that the product of two odd numbers is odd. Write them as 2a + 1 and 2b + 1. Their product is (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1, which is one more than an even number, so it is odd. For example, 7 × 9 = 63 = 2 × 31 + 1.
And they prove that if n is odd, then is odd. Write n = 2k + 1. Then , which is odd.
The usual mistakes
Assuming the conclusion. Starting from "a + b is 2 times a whole number" uses the thing to be proved, so the argument goes in a circle. A direct proof of "if n is odd, then is odd" starts from n = 2k + 1, not from .
Checking examples instead. 3 + 5 = 8 and 7 + 9 = 16 show the claim for those two pairs only.
Using one letter for two numbers. a = 2k + 1 and b = 2k + 1 would force a and b to be equal.
Stopping short of the definition. a + b = 2m + 2k + 2 is true, but the proof is not finished until the sum is written as 2 times a whole number.
House numbers on one side of a street
In the application below, four neighboring houses on the odd side of a street have the numbers 2n + 1, 2n + 3, 2n + 5 and 2n + 7. Their total is 8n + 16 = 8(n + 2), which proves that every such total is a multiple of 8.
Worked example: Four Neighboring Houses on the Odd Side of a Street, and a Fair Game's Claim About Their Total
Question On Maple Street the houses on the north side have the odd numbers 1, 3, 5, 7 and so on, with no number missing. A game at the street fair asks a player to pick four neighboring houses on the north side and add their numbers, and the stall holder claims that the total is always a multiple of 8. (a) Write the smallest of the four numbers as 2n + 1, where n is a whole number, find the total in terms of n, and use it to prove the stall holder's claim. (b) One player's total is 136. Which four houses did the player pick? Another player announces a total of 100. Can that total be right?
1.Write the smallest house number as 2n + 1, where n is a whole number with n ≥ 0. Neighboring houses on the north side differ by 2, so the four numbers are 2n + 1, 2n + 3, 2n + 5 and 2n + 7.
The four house numbers are 2n + 1, 2n + 3, 2n + 5 and 2n + 7. 2.Add them. The four 2n terms give 8n, and the numbers left over give 1 + 3 + 5 + 7 = 16, so the total is 8n + 16.
Four lots of 2n make 8n, and 1 + 3 + 5 + 7 = 16. 3.(a) Take out the common factor 8: the total is 8n + 16 = 8(n + 2). Since n + 2 is a whole number, the total is a multiple of 8 for every choice of four neighboring houses, which proves the claim.
(a) 8n + 16 = 8(n + 2), a multiple of 8 for every whole number n. 4.For a total of 136, solve 8(n + 2) = 136. Divide both sides by 8 to get n + 2 = 17, so n = 15 and the smallest number is 2 × 15 + 1 = 31. The houses are 31, 33, 35 and 37.
8(n + 2) = 136 gives n = 15, so the houses are 31, 33, 35 and 37. 5.(b) The player picked houses 31, 33, 35 and 37. A total of 100 cannot be right, because every total is a multiple of 8 and 100 ÷ 8 = 12.5 is not a whole number. Check: 31 + 33 + 35 + 37 = 136.
(b) Houses 31, 33, 35 and 37. A total of 100 is not a multiple of 8, so it cannot be right.
Answer: (a) The total is 8n + 16 = 8(n + 2), a multiple of 8 for every whole number n; (b) houses 31, 33, 35 and 37; a total of 100 cannot be right, since 100 is not a multiple of 8
Common mistakes
- Testing a few groups, such as 1 + 3 + 5 + 7 = 16 and 3 + 5 + 7 + 9 = 24, and calling the claim proved. Examples show the claim for those groups only; the total 8(n + 2) covers every group at once.
- Writing the four numbers as n, n + 1, n + 2 and n + 3. Those are four consecutive whole numbers, but neighboring houses on one side differ by 2, so their total is not 4n + 6.