Summary of topics so far..

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Summary of topics so far..

I Propositions, Predicates, Quantifiers.

I Using the above to express statements mathematically.

I Rules of inference.

I Using rules of inference to deduce answers.

I Ready to prove interesting results.

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Valid arguments vs true conclusions

Valid argument

If 2+3 = 7, then apples are blue. 2+3 = 7.

∴apples are blue.

True conclusion

If 3 is prime then 20+20 = 50. 20+20 = 50.

∴3 is prime.

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Valid arguments vs true conclusions

Valid argument

If 2+3 = 7, then apples are blue. 2+3 = 7.

∴apples are blue.

True conclusion

If 3 is prime then 20+20 = 50.

20+20 = 50.

∴3 is prime.

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Proofs

“You don’t have to believe in God, but you should believe in

The Book.” – Paul Erdos.

Proof Techniques.

I direct proofs.

I indirect proofs.

I proof by contradiction.

I contrapositive.

I proof by cases.

I existential proofs.

I forward backward reasoning.

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Proofs

“You don’t have to believe in God, but you should believe in

The Book.” – Paul Erdos.

Proof Techniques.

I direct proofs.

I indirect proofs.

I proof by contradiction.

I contrapositive.

I proof by cases.

I existential proofs.

I forward backward reasoning.

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Direct proof – examples

1. Ifn is an odd integer, then n² is odd.

2. The sum of any two rational numbers is rational.

3. For all integersa,b,c, if a|b andb|c thena|c.

4. For all integersa,b ifa|b and b|a, then a=b.

5. The square of any odd integer is of the form 8m+ 1 for some integerm.

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Direct proof – examples

1. Ifn is an odd integer, then n² is odd.

an odd integer can be written as 2k+ 1 for some integerk. 2. The sum of any two rational numbers is rational.

a rational number can be written as ^a_b,b 6= 0 . 3. For all integersa,b,c, if a|b andb|c thena|c.

if a|b, then b=ka for some integerk.

4. For all integersa,b if a|b andb|a, thena=b.

5. The square of any odd integer is of the form 8m+ 1 for some integerm.

any odd integer can be written as 4k+ 1 or 4k+ 3 for some integerk.

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Indirect proofs – examples

1. For any integern, ifn² is odd, thenn is odd.

2. √

2 is irrational.

3. No integer can be both even and odd.

4. There are infinitely many prime numbers.

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How about these?

1. For any integern,f(n) =n²−n+ 41 is prime.

2. Any even integer greater than two can be written as a sum of 2 primes.

3. a⁴+b⁴+c⁴ =d⁴ has no solutions for positive integers a,b,c,d.

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How about these?

Goldbach’s conjecture.

Euler’s conjecture – open for 218 years!

disproved by a counter example

95800⁴+ 217519⁴+ 414560⁴= 422481⁴.

Examples are good for intuition, but cannot serve as a proof!

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How about these?

Goldbach’s conjecture.

Euler’s conjecture – open for 218 years!

disproved by a counter example

95800⁴+ 217519⁴+ 414560⁴= 422481⁴.

Examples are good for intuition, but cannot serve as a proof!

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Some more claims to be verified

I The only consecutive integers less than 100 that are perfect powers are 8 and 9.

I The final decimal digit of the square of any integer is one of {0,1,4,5,6,9}.

I For positive real numbers x andy, ^(x+y₂ ⁾ ≥√ xy.

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Game of marbles

2 players play a game of marbles. There are 15 marbles to begin with and the players alternatively remove marbles. At each round a player can remove one, two, or three marbles. The person to remove the last marble wins the game.

Can player-1ALWAYS win?

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Grid and dominoes

Can you cover the grid with 2×1 tiles?

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Grid and dominoes

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Grid and dominoes

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Grid and dominoes

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Game of Chomp

Can player-1 win? ¹

1all images – courtsey google images

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Game of Chomp

I Lets start simple.

I What if it was a single strip?

I What if it was a square grid?

I Explicit strategies exist for above cases.

I General case of m×n grid.

I Every finite two player complete information game which does not end in a draw has a winning strategy for one of the players.

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Game of Chomp

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Game of Chomp

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Game of Chomp

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Game of Chomp

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Chomp revisited

Proof of existence of winning strategy for player-1.

I winning strategy for either player exists.

I player-1 can make a move which when coupled with player-2’s any move chomps a rectangular portion.

Take-aways from chomp, chess board puzzles, marbles game..

I proofs serve as a problem-solving tool.

I explicit strategy in marbles game.

I non-existence proof in chess board puzzles.

I existence proof in chomp (no explicit winning strategy)

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Chomp revisited

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Chomp revisited

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Chomp revisited

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Non-constructive proof

There exist irrational numbersx andy such thatx^y is rational.

Proof:

I

√2 is irrational – as proved in class.

I let x=√

2, y=√ 2.

I consider x^y; either it is a rational number and we are done.

I elsez =x^y is an irrational number.

I now, consider z^x.

z^x = (

√ 2

√ 2)

√ 2=

√ 2²= 2. Therefore either x^y is rational or (x^y)^x is rational.

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Non-constructive proof

Proof:

I

I let x=√

2,y =√ 2.

I now, consider z^x.

z^x = (

√ 2

√ 2)

√ 2=

√ 2²= 2. Therefore either x^y is rational or (x^y)^x is rational.

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Non-constructive proof

Proof:

I

I let x=√

2,y =√ 2.

I now, considerz^x.

z^x = (

√ 2

√ 2)

√ 2=

√ 2²= 2.

Therefore either x^y is rational or (x^y)^x is rational.

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Non-constructive proof

Proof:

I

I let x=√

2,y =√ 2.

I now, considerz^x.

z^x = (

√ 2

√ 2)

√ 2=

√ 2²= 2.

Therefore either x^y is rational or (x^y)^x is rational.

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Contradiction and contrapositive revisited

∀x P(x)→Q(x)

By contradiction

I take negation.

∃x (P(x)∧ ¬Q(x)).

I some derivations.

I derive ¬P(x).

By contrapositive

I x is an arbitrary element. assume ¬Q(x).

I some derivations.

I derive ¬P(x).

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Contradiction and contrapositive revisited

∀x P(x)→Q(x)

By contradiction

I take negation.

∃x (P(x)∧ ¬Q(x)).

I some derivations.

I derive ¬P(x).

By contrapositive

I some derivations.

I derive ¬P(x).

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Contradiction and contrapositive revisited

∀x P(x)→Q(x)

By contradiction

I take negation.

∃x (P(x)∧ ¬Q(x)).

I some derivations.

I derive ¬P(x).

By contrapositive

I some derivations.

I derive ¬P(x).

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Contradiction and contrapositive revisited

∀x P(x)→Q(x)

By contradiction

I take negation.

∃x (P(x)∧ ¬Q(x)).

I some derivations.

I derive ¬P(x).

By contrapositive

I x is an arbitrary element.

assume ¬Q(x).

I some derivations.

I derive ¬P(x).

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Contradiction and contrapositive revisited

∀x P(x)→Q(x)

By contradiction

I take negation.

∃x (P(x)∧ ¬Q(x)).

I some derivations.

I derive ¬P(x).

By contrapositive

I x is an arbitrary element.

assume ¬Q(x).

I some derivations.

I derive ¬P(x).

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Some common mistakes..

Theorem: Every integer is rational.

Proof: (by contradiction) 1. Assume not.

2. Then every integer is irrational.

3. 1 is an integer and therefore 1 is irrational. 4. But 1 = ¹₁ which means that 1 is rational.

5. This is a contradiction, therefore our assumption is false. 6. The theorem is true.

Mistake: Incorrect negation.

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Some common mistakes..

3. 1 is an integer and therefore 1 is irrational.

4. But 1 = ¹₁ which means that 1 is rational.

5. This is a contradiction, therefore our assumption is false.

6. The theorem is true.

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Some common mistakes..

3. 1 is an integer and therefore 1 is irrational.

4. But 1 = ¹₁ which means that 1 is rational.

5. This is a contradiction, therefore our assumption is false.

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Some common mistakes..

Theorem: Product of any two odd integers is odd.

Proof:

1. Assume m andn are odd integers.

2. Ifmn is odd then mn= 2k+ 1 for some integerk. 3. m= 2a+ 1 for some integera.

4. n = 2b+ 1 for some integerb.

5. mn = (2a+ 1)(2b+ 1) = 2k+ 1. from step-2. 6. Therefore mn is odd.

Mistake: Begging the question.

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Some common mistakes..

Proof:

5. mn = (2a+ 1)(2b+ 1) = 2k+ 1. from step-2.

6. Therefore mn is odd.

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Some common mistakes..

Proof:

5. mn = (2a+ 1)(2b+ 1) = 2k+ 1. from step-2.

6. Therefore mn is odd.

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Some common mistakes..

Theorem: Difference between an odd and even integer is odd.

Proof:

1. Assume m is odd and n is even. 2. m= 2k+ 1.

3. n = 2k.

4. m−n= 2k+ 1−2k = 1.

5. 1 is odd and therefore the theorem is true.

Mistake: Using the same variable in steps 2 and 3.

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Some common mistakes..

Proof:

1. Assume m is odd andn is even.

2. m= 2k+ 1.

3. n = 2k.

4. m−n= 2k+ 1−2k = 1.

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Some common mistakes..

Proof:

1. Assume m is odd andn is even.

2. m= 2k+ 1.

3. n = 2k.

4. m−n= 2k+ 1−2k = 1.

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To summarize..

Ingredients of a good proof

I A clever idea!! heart of the proof.

I A step-wise clear and logical presentation.

I Avoid “it is obvious” – instead give a short justification.

I If needed, split the proof into lemmas/ smaller claims.

Writing good proofs is an art.