Download Proof Techniques for Mathematical Statements and more Lecture notes Logic in PDF only on Docsity!
Proof Techniques
Jessica Su
November 12, 2016
1 Proof techniques
Here we will learn to prove universal mathematical statements, like “the square of any odd number is odd”. It’s easy enough to show that this is true in specific cases – for example, 32 = 9, which is an odd number, and 5^2 = 25, which is another odd number. However, to prove the statement, we must show that it works for all odd numbers, which is hard because you can’t try every single one of them.
Note that if we want to disprove a universal statement, we only need to find one counterex- ample. For instance, if we want to disprove the statement “the square of any odd number is even”, it suffices to provide a specific example of an odd number whose square is not even. (For instance, 3^2 = 9, which is not an even number.)
Rule of thumb:
- To prove a universal statement, you must show it works in all cases.
- To disprove a universal statement, it suffices to find one counterexample.
(For “existence” statements, this is reversed. For example, if your statement is “there exists at least one odd number whose square is odd, then proving the statement just requires saying 32 = 9, while disproving the statement would require showing that none of the odd numbers have squares that are odd.)
1.0.1 Proving something is true for all members of a group
If we want to prove something is true for all odd numbers (for example, that the square of any odd number is odd), we can pick an arbitrary odd number x, and try to prove the statement for that number. In the proof, we cannot assume anything about x other than that it’s an odd number. (So we can’t just set x to be a specific number, like 3, because then our proof might rely on special properties of the number 3 that don’t generalize to all odd numbers).
Example: Prove that the square of any odd number is odd.
1.1 Special techniques 1 PROOF TECHNIQUES
Proof: Let x be an arbitrary odd number. By definition, an odd number is an integer that can be written in the form 2k + 1, for some integer k. This means we can write x = 2k + 1, where k is some integer. So x^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1. Since k is an integer, 2k^2 + 2k is also an integer, so we can write x^2 = 2+ 1, where = 2k^2 + 2k is an integer. Therefore, x^2 is odd.
Since this logic works for any odd number x, we have shown that the square of any odd number is odd.
1.1 Special techniques
In addition to the “pick an arbitrary element” trick, here are several other techniques com- monly seen in proofs.
1.1.1 Proof by contrapositive
Consider the statement “If it is raining today, then I do not go to class.”
This is logically equivalent to the statement “If I go to class, then it is not raining today.”
So if we want to prove the first statement, it suffices to prove the second statement (which is called the contrapositive).
Note that it is not equivalent to the statement “If I do not go to class, then it is raining today” (this is called the fallacy of the converse).
Example: Let x be an integer. Prove that x^2 is an odd number if and only if x is an odd number.
Proof: The “if and only if” in this statement requires us to prove both directions of the implication. First, we must prove that if x is an odd number, then x^2 is an odd number. Then we should prove that if x^2 is an odd number, then x is an odd number.
We have already proven the first statement, so now we just need to prove the second state- ment. The second statement is logically equivalent to its contrapositive, so it suffices to prove that “if x is an even number, then x^2 is even.”
Suppose x is an even number. This means we can write x = 2k for some integer k. This means x^2 = 4k^2 = 2(2k^2 ). Since k is an integer, 2k^2 is also an integer, so we can write x^2 = 2for the integer = 2k^2. By definition, this means x^2 is an even number.
1.1.2 Proof by contradiction
In proof by contradiction, you assume your statement is not true, and then derive a con- tradiction. This is really a special case of proof by contrapositive (where your “if” is all of mathematics, and your “then” is the statement you are trying to prove).
1.2 Proof by induction 1 PROOF TECHNIQUES
- If the statement is true for n = 1, then it is also true for n = 2
- If the statement is true for n = 2, then it is also true for n = 3
- If the statement is true for n = 3, then it is also true for n = 4 -...
Together, these implications prove the statement for all positive integer values of n. (It does not prove the statement for non-integer values of n, or values of n less than 1.)
Example: Prove that 1 + 2 + · · · + n = n(n + 1)/2 for all integers n ≥ 1.
Proof: We proceed by induction.
Base case: If n = 1, then the statement becomes 1 = 1(1 + 1)/2, which is true.
Inductive step: Suppose the statement is true for n = k. This means 1 + 2 + · · · + k = k(k +1)/2. We want to show the statement is true for n = k +1, i.e. 1+2+· · ·+k +(k +1) = (k + 1)(k + 2)/2.
By the induction hypothesis (i.e. because the statement is true for n = k), we have 1 + 2 + · · · + k + (k + 1) = k(k + 1)/2 + (k + 1). This equals (k + 1)(k/2 + 1), which is equal to (k + 1)(k + 2)/2. This proves the inductive step.
Therefore, the statement is true for all integers n ≥ 1.
1.2.1 Strong induction
Strong induction is a useful variant of induction. Here, the inductive step is changed to
- Base case: The statement is true when n = 1.
- Inductive step: If the statement is true for all values of 1 ≤ n < k, then the statement is also true for n = k.
This also produces an infinite chain of implications:
- The statement is true for n = 1
- If the statement is true for n = 1, then it is true for n = 2
- If the statement is true for both n = 1 and n = 2, then it is true for n = 3
- If the statement is true for n = 1, n = 2, and n = 3, then it is true for n = 4 -...
Strong induction works on the same principle as weak induction, but is generally easier to prove theorems with.
Example: Prove that every integer n greater than or equal to 2 can be factored into prime numbers.
2 REFERENCES
Proof: We proceed by (strong) induction.
Base case: If n = 2, then n is a prime number, and its factorization is itself.
Inductive step: Suppose k is some integer larger than 2, and assume the statement is true for all numbers n < k. Then there are two cases:
Case 1: k is prime. Then its prime factorization is just k.
Case 2: k is composite. This means it can be decomposed into a product xy, where x and y are both greater than 1 and less than k. Since x and y are both less than k, both x and y can be factored into prime numbers (by the inductive hypothesis). That is, x = p 1... ps and y = q 1... qt where p 1 ,... , ps and q 1 ,... , qt are prime numbers.
Thus, k can be written as (p 1... ps) · (q 1... qt), which is a factorization into prime numbers.
This proves the statement.
2 References
Greg Baker, “Introduction to Proofs”
https://www.cs.sfu.ca/~ggbaker/zju/math/proof.html
CS 103 Winter 2016, “Guide to Proofs”
http://stanford.io/2dexnf
Peng Hui How, “Proof? A Supplementary Note For CS161”
http://web.stanford.edu/class/archive/cs/cs161/cs161.1168/HowToWriteCorrectnessProof.pdf
