CSCI 2824: Lecture Notes # 6

In these 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.

Basic Proof Strategies

How to prove a theorem of course depends on what you are asked to prove. We will give you some templates for how a proof must proceed. Of course this works only for simple theorems. For complex theorems, the idea is to decompose into simpler claims. Proving the simpler claims, we build upon them to prove more complex claims and so on.

We will restrict ourselves to facts about numbers for now.

Universal Statements

For every number n, some fact holds over n: (forall n in mathbb{Z}) P(n)

Proofs of Universal Statements

Universal statements are proved as follows:

  • Assume an arbitrary number n.

  • Prove that P(n) will hold usually by doing some algebra.

Let us now look at an example.

Example #1

Theorem For every natural number n, the number  2 n +1 is an odd number.

  • Let n be any given number and let m= 2n+1.

  • We are asked to show that m is odd.

  • We will show therefore, that m leaves a remainder of 1 when divided by 2.

 m mod 2 = (2 n + 1) mod 2 = (2 n mod 2) + (1 mod 2) = 0 + 1 = 1


Universal Statements With Implication

We now look at a special form of universal statements of the form:

Universal Statement With Implication

 (forall n in mathbb{N}) P(n) Rightarrow Q(n).

Following the general rule for universal statements, we write a proof as follows:

  • Let n be any given number in mathbb{N}.

  • There are two cases P(n) does not hold or P(n) holds.

  • In case P(n) does not hold, the implication trivially holds.

  • In case P(n) holds, we will now prove Q(n).

    • Typically, some algebra here to show that P(n) Rightarrow Q(n).

We can use a simple short-cut that avoids unnecessary language in such proofs.

  1. Let n be any given number in mathbb{N} such that P(n) holds.

  2. We will show that under these assumptions Q(n) holds.

    • Typically, some algebra to show that Q(n) holds.

Example #1

Theorem If n is an even number and n > 2 then n is composite.


Here are our reasoning steps:

  • Assume a given number n such that n is even and n > 2.

  • From this assumption, we can write n = 2m wherein m > 1.

    • n has two non-trivial factors 2 and m.

    • n is therefore composite. QED.

Example #2

Theorem For every natural number n, If n geq 3 then n^2 -1 cannot be prime.


Here are the steps of our reasoning.

  • Assume a given number n such that n geq 3.

  • We will now show that n^2 -1 is a composite number.

    • By elementary algebra, we see that (n^2 -1) = (n-1) (n+1).

    • We note that n^2-1 can be written as the product of two numbers.

    • It remains to show that the factors of n^2-1 are not trivial. That is, n-1 not=1 ( and therefore n +1 not= (n^2 -1)).

      • Since n geq 3, n-1 geq 2. Therefore n-1 not=1.

    • We have just shown that n^2-1 = (n-1) (n+1) and both n-1 is greater than 1. Therefore n^2-1 is composite. QED.

Existential Statements

We will mention existential statements. They are rarer but interesting.

Simple existential statements assert that a number with some property exists.

(exists n in mathbb{N}) P(n).

To prove an existential statement, we just give an example.


Theorem There exist two prime numbers whose sum is also a prime number.


2 and 3 serve as examples to our theorem.

Diversion: Constructive Proofs

Existential statements can be proved in another way without producing an example. Typically this involves a proof by contradiction (we will study these types of proofs soon). Such proofs are called non-constructive proofs.

Theorem There exist two irrational numbers a and b such that a^b is rational.


We will show that such numbers exist without giving you a concrete example.

Consider the number a = sqrt{2}^{sqrt{2}} and b = sqrt{2}.

 a^b = (sqrt{2}^{sqrt{2}})^{sqrt{2}} = sqrt{2}^{(sqrt{2}cdot sqrt{2})} = sqrt{2}^2 = 2

Therefore a^b is rational. We know that b is irrational.

There are two cases:

  1. a is irrational. In that case, we are done since a,b are both irrational and a^b is rational.

  2. a is rational.

    1. In that case, a = sqrt{2}^{sqrt{2}} is rational.

    2. But sqrt{2} is irrational itself.

    3. So in this case, we have c=d = sqrt{2} are irrational and c^d = a is rational by assumption.

As a result, there must exist two numbers a,b such that a^b is rational while a,b themselves are irrational. ( Our argument just has not produced any concrete example to point to. :-) )


Discuss some of the philosophical implications of constructivism vs. non-constructivism

Needless to say we will leave non-constructive proofs to mathematicians and the debate to the philosophers for now. If you are interested, these ideas are usually covered in a philosophy of mathematics or a philosophy of science class.

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 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 prime.

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”?


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?


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.


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: 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??


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.