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Proofs — basic strategies for proving universal statements (CSCI 2824, Spring 2015)

In this series of notes, we are going to

  1. Provide proofs of various propositions

  2. Ask you to fix some of our proofs.

  3. Practice proving theorems by expressing your argument in a succinct and logically consistent form.

It is often enlightening to learn from mistakes. On this page, we study examples of wrong proofs (and hopefully learn to not repeat them in homework/exams!).

Flawed Mathematical Arguments

We will now see examples of flawed arguments that you need to watch out for when doing mathematics. Examples include

  • Proving by example: Just present a few examples and note that an universal statement holds based on these.

  • Assuming some fact in the proof that does not follow from the premise.

  • Proving by intuition: Appeal to your intuition usually by drawing a diagram.

  • Statements that should never be present in a good proof:

    • it is clear to me (proof by vision!! )

    • it just feels right (the touchy-feely proof!! )

    • intuitively,… (proof by intuition!! )

    • my professor told us that,… (proof by authority!! )

    • it is obvious,.. (unless it truly truly truly is rather obvious as 1+1 = 2 !!).

  • Proving the wrong statement.

    • Proof of the converse (instead of proving P Rightarrow Q, we prove Q Rightarrow P).

    • Proof of something unrelated.

  • Not proving the statement for all the cases.

    • Missing out on some corner cases.

      • Proof only applies to n even but not to n odd.

Example# 1

Claim: For n > 1, if n is even, then n^2+1 is prime. I.e, forall n in mathbb{N} (n !! mod 2 = 0) Rightarrow n^2+1 mbox{ is prime}.

Proof Attempt # 1

Let us test for n=2, we have n^2+1 is 5. Works. It also works for n=4 since n^2+1 = 17 is prime and 6 since 6^2+1 is prime. Therefore, n^2 +1 is prime if n is even.

Let us attempt one more proof of this:

Proof Attempt # 2

Assume n^2+1 is prime. We will prove that n must be even.

  • Since n^2+1 is prime, and n > 1, n^2+1 must necessarily be odd.

  • Since n^2+1 is odd, it must be the case that n^2 is even.

  • If n^2 is even, we have proved previously that n is also even.

  • Therefore, n is even. QED??

Are there any flaws in either of these proofs? Do they convince you of the truth of our “claim”?

Answer

The claim is false in the first place because it fails for n=8, wherein 8^2+1 = 65 = 13 times 5.

The first proof attempt is a proof by example which is generally invalid for universally quantified statements.

The second proof attempt actually sets out to prove the converse. Instead of proving n^2+1 is prime, it assumes this and tries to prove, instead, that n is even.

Example #2

Claim If two numbers a and b are odd, then a+b is even.

Exercise: Write this down in logical notation.

Let us look at a proof:

Attempted Proof

Proof Here are our reasoning steps:

  1. Since a is odd, it can be written as 2 n +1 for some n.

  2. Since b is odd, it can be written as 2 n+1 too for some n.

  3. Therefore a+b = 2n +1 + 2n + 1 = 4 n +2 .

  4. But 4n +2 = 2 ( 2 n +1) is an even number.

  5. Therefore, a+b is even. QED.

Is there anything wrong with the proof above?

Now let us look at a related claim:

Claim-2 If two numbers a and b are odd, then a+b mod 4 = 2.

Is this a true statement?

Proof

Here are our reasoning steps:

  1. Since a is odd, it can be written as 2 n +1 for some n.

  2. Since b is odd, it can be written as 2 n+1 too.

  3. Therefore a+b = 2n +1 + 2n + 1 = 4 n +2 .

  4. a + b mod 4 = 4n +2 mod 4 = 2.

  5. Therefore, a+b mod 4 = 2. QED.

Can you correct the demonstrations above? What went wrong.

Answer

The problem was in assuming that b = 2n +1 for some n. By saying that a = 2n+1, for some n and b = 2n+1 for some n, there is a flawed assumption that a = b = 2n+1, which was never warranted.

Therefore, we are able to “prove” Claim-2, which is clearly false. For example, a=3 and b=5 yields us a+b = 8 and 8 mod 4 = 0.

Claim-1 is correct and the corrected proof is as follows:

Claim-1 If two numbers a and b are odd, then a+b is even.

Corect Proof

Proof Here are our reasoning steps:

  1. Since a is odd, it can be written as 2 n +1 for some n.

  2. Since b is odd, it can be written as 2 mathbf{k}+1 too for some k.

  3. Therefore a+b = 2n +1 + 2k + 1 = 2(n+k) +2 .

  4. But 2(n+k) +2 = 2 ( n+k +1) = 2 l is an even number.

  5. Therefore, a+b is even. QED.

Example #3

Claim If n is natural number then n^2-1 is a composite number.

Proof

Proof: Let n be a natural number.

  1. We can write n^2 -1 as a product of (n+1) (n-1).

  2. Therefore n^2 -1 is a composite number. QED??

Answer

The claim is actually false. Take n = 2, we have 2^2 - 1 = 3, a prime number.

What went wrong in the proof? Well, we are correct in writing as n^2 -1 = (n+1) (n-1) but this does not immediately show that n^2-1 is composite. We have to convince ourselves that n+1 not=1 and n-1 not = 1. Recall:

Definition: Prime and Composite Numbers

A natural number n geq 2 is composite if it can be written as n = m times p for natural numbers m,p where m,p cannot be 1 or n itself. In logic, we define a predicate mbox{isComposite}(n) as follows:

(n geq 2) mbox{AND} (exists m,p in mathbb{N} ( m times p = n mbox{AND} m not= 1 mbox{AND} p not= 1),..

Likewise, natural number n geq 2 is prime if n = m times p for some natural numbers m,p, then m= 1 or p= 1. In logic, we define a predicate mbox{isPrime} for natural numbers, as follows:

mbox{isPrime}(n): (n geq 2) mbox{AND} (forall m,p inmathbb{N} ( m times p = n) Rightarrow ( m = 1 mbox{OR} p = 1) )

An important exception involves the numbers 0,1. These are taken to be neither prime nor composite.

The proof above can only be correct when n not = 0 and n not = 2.