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

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

Example# 1

Claim: For n >1, if nis even, then n^2+1is 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+1is 5. Works. It also works for n=4since n^2+1 = 17is prime and 6since 6^2+1is prime. Therefore, n^2 +1is prime if nis even.

Let us attempt one more proof of this:

Proof Attempt # 2

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

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+1is prime, it assumes this and tries to prove, instead, that nis even.

Example #2

Claim If two numbers aand bare odd, then a+bis even.

Exercise: Write this down in logical notation.

Let us look at a proof:

Attempted Proof

Proof Here are our reasoning steps:

  1. Since ais odd, it can be written as 2 n +1for some n.
  2. Since bis odd, it can be written as 2 n+1too 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+bis even. QED.

Is there anything wrong with the proof above?

Now let us look at a related claim:

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

Is this a true statement?

Here are our reasoning steps:

  1. Since ais odd, it can be written as 2 n +1for some n.
  2. Since bis odd, it can be written as 2 n+1too.
  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 +1for some n. By saying that a = 2n+1, for some nand b = 2n+1for 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=3and b=5yields us a+b = 8and 8 mod 4 = 0.

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

Claim-1 If two numbers aand bare odd, then a+bis even.

Corect Proof

Proof Here are our reasoning steps:

  1. Since ais odd, it can be written as 2 n +1for some n.
  2. Since bis 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 lis an even number.
  5. Therefore, a+bis even. QED.

Example #3

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

n

Proof: Let be a natural number.

  1. We can write n^2 -1as a product of (n+1) (n-1).
  2. Therefore n^2 -1is 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^2-1is composite. We have to convince ourselves that n+1 not=1and n-1 not = 1. Recall:

Definition: Prime and Composite Numbers

A natural number n geq 2is composite if it can be written as n = m times pfor natural numbers m,pwhere m,pcannot be 1or nitself. In logic, we define a predicate mbox<isComposite>(n) as follows:

 (n geq 2) mbox</p>
<p> (exists m,p in mathbb ( m times p = n mbox m not= 1 mbox p not= 1),..

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

mbox<isPrime></p>
<p>(n): (n geq 2) mbox (forall m,p inmathbb ( m times p = n) Rightarrow ( m = 1 mbox p = 1) )

0,1

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

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