Proof Techniques

Stage 22 of 23 Strand 2 of 4 7 lessons

7 illustrated lessons, each teaching the why before the how.

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Direct Proof

Start at the hypothesis and walk to the claim.

A direct proof starts from the hypothesis and reaches the conclusion by steps that are each forced

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claim: the sum of two odd numbers is evenwrite the hypothesis in symbolslet a = 2m + 1 and b = 2k + 1

A direct proof begins by writing what you are given in a form you can work with.

a + b = (2m + 1) + (2k + 1)collect the terms= 2m + 2k + 2take the 2 out= 2(m + k + 1)

Each line follows from the one above by algebra, with nothing assumed on the way.

a + b = 2(m + k + 1)this is the definition of evenso a + b is even

The last line matches the definition of even, so the claim is reached and the proof ends.

the movestartassume the hypothesismiddleforced steps onlyendstate the conclusion

That is the whole method: assume the hypothesis, take forced steps, state the conclusion.

Now you

assume Pforced stepsconclude Q

What does a direct proof never do?

assume Pforced stepsconclude Q

A direct proof of "if n is odd then n² is odd" starts how?

Proof by the Contrapositive

Prove the equivalent form that you can use.

Proving the contrapositive proves the original statement, and it is worth doing when the negated form is easier to work with

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claim: if is even, then n is even
no useful factorization
starting from even gives you little

Starting from n² even leaves you nothing to write down, so a direct proof stalls here.

if is even, then n is even
the contrapositive
if n is odd, then is odd

The contrapositive of the claim is true exactly when the claim is, so prove that instead.

n = 2k + 1square it out
odd, by the definition

Now the hypothesis is usable: n odd means n = 2k + 1, and squaring shows n² is odd.

the contrapositive is provedthe two share a truth valueso the original claim is proved

Proving the contrapositive proves the claim. There is no extra step at the end.

Now you

if P then Qprove this insteadif not Q then not P

After proving the contrapositive, what remains to be done?

if P then Qprove this insteadif not Q then not P

Which claim is the contrapositive route best suited to?

Proof by Exhaustion

Cover every case, and leave no case out.

Splitting a claim into a finite list of cases proves it only when the list leaves no case out

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claim: is even for every whole n
two cases, and no third
every whole n is even or odd

Choose a way of splitting that covers everything. Every whole number is even or odd.

n = 2keven

Take the cases one at a time. With n even the expression comes out even.

n = 2k + 1even again

With n odd it comes out even as well, so both cases land on the same answer.

case even: provedand there is no other casecase odd: provedso the claim holds for every n

The proof works only because the two cases together leave no whole number out.

Now you

case A: provedcase B: provednothing left over

A claim about n mod 3 is split into n = 3k and n = 3k + 1. What is wrong?

case A: provedcase B: provednothing left over

What makes a case split a proof rather than a collection of examples?

Disproof by Counterexample

One failing case ends a claim about every case.

One case where a general claim fails disproves it, while any number of cases that fit proves nothing

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claim: is prime for every whole n

A claim about every case is a universal statement, and it allows no exceptions anywhere.

n = 0: 41, primestill fittingn = 1: 43, primeand still fittingn = 2: 47, prime

The first forty values all come out prime. That is encouraging, but it is not a proof.

n = 41:
every term carries a factor of 41
= 41(41 + 1 + 1)

At n = 41 every term carries a factor of 41, so the value is 41 × 43, which is not prime.

one failing case is enoughnothing more is neededthe claim is disproved

One failure is enough. To disprove a claim about every case, a single counterexample is all you need.

Now you

for every whole n

Which one counterexample disproves the claim drawn?

The Pigeonhole Principle

More objects than boxes forces a double.

When more objects than boxes are placed into the boxes, some box has to hold more than one

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still to deal

There are five objects and four boxes. Put one in each box and the boxes are full.

still to deal2111

The fifth object has nowhere new to go, so some box ends up holding two.

n objects into k boxes, with n > khowever you arrange themsome box holds at least two

How you arrange them never matters. More objects than boxes forces some box to hold at least two.

a fraction with denominator 7so one repeatsonly 6 nonzero remainders existthe decimal recurs

A long division has only so many remainders, so one must repeat and the decimal recurs.

Now you

still to deal

Put 8 objects into 3 boxes. What is the largest number some box must hold?

still to deal

A drawer holds socks in 6 colors, mixed up in the dark. How many must you take to be sure of a matching pair?

Proving Existence and Uniqueness

Build one, then force any two to coincide.

Existence is proved by producing one such object, and uniqueness by forcing any two of them to be equal

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claim: some whole n has
one is exhibited
n = 2 gives 4 − 10 + 6 = 0

An existence claim is settled by producing one object and checking it. Nothing more is needed.

claim: 2x + 3 = 11 has a solutionexistence, by constructionx = 4 works

Building the object is the usual way, and the construction is itself the proof.

suppose 2a + 3 = 11 and 2b + 3 = 11subtract 3 from each2a = 2bso there was only ever onea = b

For uniqueness, take any two that work and force them to be the same object.

what you showexistenceproduce oneuniquenessany two coincide

Existence and uniqueness are separate claims. Producing one object says nothing about whether a second exists.

Now you

what you showexistenceproduce oneuniquenessany two coincide

How is "there is a whole number whose square is 49" proved?

what you showexistenceproduce oneuniquenessany two coincide

What does the phrase "there exists a unique n" claim?

Counting by a Bijection

Match a hard count to an easy one, exactly.

Matching a hard collection one to one with an easy one proves the two have the same size

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how many ways to put 7 identical coins into 3 jars?the coins are indistinguishablejars may be left empty

The ways of filling the jars are hard to count directly, because the coins are all alike.

*×*×|×*×*×*×|×*×*jars of 2, 3 and 2

Write the coins as stars and the walls between jars as bars, in one row.

7 stars and 2 bars in a rowand each way gives one roweach row is one way of filling the jars

Every row of stars and bars matches exactly one filling, and every filling matches exactly one row.

choose which 2 of the 9 places hold barsthe easy count, so the hard one is 36 too9C2 = 36

Counting the rows means choosing which 2 of the 9 places hold bars, so the answer is 36.

Now you

?×?×?×?×?×?×?×?×?×?3 of these places hold bars

How many ways can 7 identical coins go into 4 labeled jars, empties allowed?

?×?×?×?×?×?×?×?×?2 of these places hold bars

How many ways can 7 identical coins go into 3 labeled jars, empties allowed?

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