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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.

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.

Three types of proof strategies

Over the next 6 lectures or so, we will cover Chapter 2 of the textbook and learn the following three types of proof strategies:

  • Direct proof

  • (Strong and weak) mathematical induction

  • Proof by contradiction

In general, some good rules of thumb include the following.

  • Be organized when writing down your arguments.

    • (If you find it hard to follow your arguments, then you can't expect the reader to be able to follow, right?)

  • Start a new line for each argument step. (This often leads to better organization.)

  • Don't skip major argument steps (even if it seems easy to do).

We will restrict ourselves to facts about numbers for now.

Universal Statements

A universal statement (over a certain set a.k.a ‘‘universe of discourse’’ D) is a claim that for every number n in D, some fact (described by some predicate) holds over n. Mathematically, a universal statement is in the form (forall n in D) P(n).

Proofs of Universal Statements

Universal statements (over a set D) are proved as follows:

  • Fix an arbitrary element nin D.

  • Prove that P(n) holds — usually by doing some algebra.

Let us now look at an example.

Example 1

Let's try to prove the following theorem.

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

Idea behind the proof construction

  • Let n be any fixed natural number.

  • Let m= 2n+1.

  • We are asked to show that m is odd.

  • What does it mean for a natural number to be odd? By definition, that number has to leave a remainder of 1 when divided by 2.

  • Let's see if m has a remainder 1 when divided by 2:

begin{array}{rcl}hspace{10em} mhspace{-0.6em} mod 2 &=& (2 n + 1)hspace{-0.6em} mod 2 &=& (2 nhspace{-0.6em} mod 2) + (1hspace{-0.6em} mod 2) &=& 0 + 1 &=& 1 end{array}

  • We can conclude that m is an odd number.

The actual proof

Let n be any fixed natural number. Let m=2n+1. Then

begin{array}{rcl}hspace{10em} mhspace{-0.6em} mod 2 &=& (2 n + 1)hspace{-0.6em} mod 2 &=& (2 nhspace{-0.6em} mod 2) + (1hspace{-0.6em} mod 2) &=& 0 + 1 &=& 1 end{array}

Hence m=2n+1 is an odd number. QED.

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} bigl(P(n) Rightarrow Q(n)bigr).

We recall that for given two propositions p and q (or predicates), the implication pimplies q is true if one of the following holds:

  • the premise p and the conclusion q are both true;

  • the premise p is false. (In this case it does not matter whether the conclusion holds true or not.)

(An easy example is “If you win the bet, then I will give you $20.” If you don't win the bet, I can still give you $20 without breaking my promise!)

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

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

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

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

  • In the case where 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 fixed number in mathbb{N} such that P(n) holds.

  2. We will show that under these assumptions Q(n) holds (typically by using some algebra).

Example 2

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

Proof

Here are our reasoning steps:

  • Fix any natural 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 3

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

Proof

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.

Now we are going to learn two commonly used techniques: proof by cases and proof by contrapositive (for implication statements).

Proof by contrapositive

This technique is used for proving implications of the form P(x)implies Q(x). Since an implication is always equivalent to its contrapositive, proving that neg Q(x)implies neg P(x) does the job.

Example 4

Theorem. For any integer n, if n^2 is even, then n is even.

The statement in this theorem is equivalent to the contrapositive statement: For any integer n, if n is not even, then n^2 is not even. It suffices to prove this contrapositive statement here!

Proof

  • Fix any odd integer n.

  • By definition of odd integers, n=2m+1 for some integer m.

  • Then n^2 = (2m+1)^2 = 4m^2+4m+1 = 2(m^2+2m)+1.

  • Hence n^2 is an odd integer. QED.

Caution: This technique is often useful but make sure you formulate the contrapositive statement properly before proving it!

Proof by cases

When given a statement to prove, sometimes it is easier to consider severl complementary scenarios, and prove the statement in each of the scenarios via different arguments.

Example 5

Prove that for any positive integers x,y, if x^2+y^2geq9, then x+ygeq 3.

Intuition

Since we are only concerned with positive integers 1,2,3,ldots and the conclusion x+ygeq 3 automatically holds when x or y is greater than or equal to 3, we can consider the following cases:

  • Case 1: xgeq 3 or ygeq 3.

  • Case 2: neg bigl((xgeq 3) lor (ygeq 3)bigr), or equivalently, xleq 2 and yleq 2.

Formal proof

For any positive integers x,y, we consider two different cases.

  • Case 1: xgeq 3 or ygeq 3. Without loss of generality, suppose that xgeq 3. Then the conclusion holds because x+ygeq x geq 3.

  • Case 2: xleq 2 and yleq 2. We check if the premise x^2+y^2 could ever hold in this case — there are only 3 cases to check because x,y are positive integers!

    • If x=1 and y=1, then x^2+y^2=2<9 so the premise does not hold.

    • If x=2 and y=1 (or symmetrically x=1 and y=2), then x^2+y^2=5<9. So again the premise does not hold.

    • If x=2 and y=2, then x^2+y^2=8<9 so the premise does not hold. Therefore in Case 2, the premise x^2+y^2geq9 never holds. Then the implication (x^2+y^2geq 9 ) implies (x+ygeq3) trivially holds in this case.

In conclusion, the implication (x^2+y^2geq 9 ) implies (x+ygeq3) holds for all positive integers x,y.

QED

Universal statements involving ‘‘if and only iff’’

Finally, we look at the general proving techniques for one common type of universal statements:

Universal statements with implication and converse

forall, xin D (P(x)iff Q(x))

For proving such a statement, we usually proceed in two main steps after fixing any xin D.

  • Prove that P(x) implies Q(x).

  • Prove that Q(x) implies P(x).

Example 6

Theorem. For any integer n, n is even if and only if n^2 is even.