Lecture 2: Proofs

Proof by Contraposition

"If x then y" is logically equivalent to "If not y, then not x" the second version is the contrapositive of the first

opposite of "if X, then Y"

"X AND not Y", because if X is false, then the if-then statement is true . only way to make an if-then statement false is to say that "if X is true AND the then is false"

opposite of "there exists" statements

"for all" statements

opposite of "n has both property X and Y"

"n either doesn't have property X or doesn't have property Y (or neither X nor Y"

opposite of "for all" statements

"there exists" symbol (backward E)

Claim: If N is an odd integer, then N^2 is an odd integer

- direct proof: start with given and turn into assertion being made - n is an odd integer, so for some k - n= 2k+1 - n^2 = 4k^2 + 4k + 1 - n^2 = 2* (2k^2 + 2k) + 1 - the +1 makes (2k^2+2k) odd - therefore, n^2 is odd

Claim: if R and S are rational numbers, then R+S is rational

- r and s are rational numbers so - r = m/n , s = p/q - add r and s together with same denominator - r + s = (mp+pn)/nq - therefore, rational

direct proofs steps

- start with assumptions- aka premises- and start deriving facts till you get conclusion - just keep deriving things - no one knows how a proof will take form till they've found it - once you derive conclusion, clean up work by removing unneeded things

Proof by Contradiction general example

- want to prove X - Assume not X - Derive Y and not Y - Contradiction! Therefore, X

Proof by Contradiction definition

assume opposite of what you're trying to prove -> derive a logical impossibility if you have logical impossibility, then it means that there's an error. If you do the math correctly, means that only error was with the assumption. so logical opposite of assumption must be correct

Prove by contraposition based off of prove directly: for all (upside down A) N, N is odd if and only if (IFF) N^2 is odd.

2) If N^2 is odd, then N is odd Contraposition: If N is even, then N^2 is even -> easier to show n = 2k n^2 = 4k^2 n^2 = 3 * (2k^2) Therefore, n is odd IFF n^2 is odd

Prove DIRECTLY: for all (upside-down A) N, if 3N+2 is odd, then N is odd.

3n+2=2n+2 3n+2 is odd 2n+2 is even odd - even = odd 3n + 2 - 2n + 2 = n

Proof by Contradiction: for all (upside-down A) N, if 3N+2 is odd, then N is odd.

Proof by Contradiction: opposite of an 'If then' statement is a 'AND' statement -> Assume 3n+2 is odd AND N is even n = 2k 3n = 6k 3n + 2 = 6k + 2 3n + 2 = 2* (3k+1) 3n + 2 is even, but we assumed it was odd. Contradiction!

Where does DeMorgan's law apply for the contrapositive of for all (upside down A) positive integers A and B, if N = A*B, then A <= √(n) or B<= √(n)

DeMorgan's law applies for "if A > √n AND b> √n" b/c it applies to both of these if statements

Prove directly: for all (upside down A) N, N is odd if and only if (IFF) N^2 is odd.

IFF makes it so we have 2 proofs to do before we can do IFF 1) If N is odd, then N^2 is odd n=2k+1 n^2= 2(2k^2 + 2k) + 1 2) If N^2 is odd, then N is odd n^2 = 2k + 1 n = √(2k+1) ...? therefore, need to prove by contraposition

Prove √2 is irrational by contradiction trial 1

Prove by Contradiction: Assume √2 is rational so √2 = p/q for some integers p and q 2 = p^2/ q^2 2q^2 = p^2 -> so p^2 is even -> p = 2k 2 = 4k ^2/q^2 q^2 = 2 k^2 (getting rid of denominator) so q^2 is even, which means q is even. so q= 2r √2 = (p/q) = (2r/2k) = (r/k) therefore, r is even and k is even, so r = 2s, k = 2t. therefore, s is even and t is even... this is wrong - it'll just be constantly divided by 2

Prove √2 is irrational by contradiction trial 2

assume √2 is rational √2 = p/q, where p/q is a reduced fraction!!! THIS IS THE IMPORTANT PART -> YOU ASSUME P/Q IS ALREADY THE REDUCED FRACTION 2 = p^2/q^2 2q^2 = p^2 -> p=2k 2q^2 = 4k^2 -> q= 2r contradiction! we assumed p/q was a reduced fraction, but both p and q are divisible by 2

there exists symbol

backwards E

Prove: for all (upside down A) positive integers A and B, if N = A*B, then A <= √(n) or B<= √(n)

direct proof will be a challenge prove by contraposition if not Y, then not X if A > √n AND b> √n, then N does NOT = a*b - a> √(n) and b > √(n) a*b > √(n)√(n) - a * b does NOT equal √n

Claim: disprove that all primes are odd

disprove a general statement using counterexample. backwards E ("there exists") a even prime, 2

example of a proof by contraposition irl

doesnt make sense if you say: if it's cloudy, it's raining -> not entirely accurate true statement: if it's raining, then it's cloudy contraposition: if it's not cloudy, it's not raining

negation of "there exists (backwards E) an integer n with property X"

for all (upside down A) integers n, n doesn't have property X

if and only if - for all (upside down A) N, N is odd if and only if (IFF) N^2 is odd.

goes in both directions If N is odd then N^2 is odd if N^2 is odd, then N is odd left implies right and right implies left

one of his favorite things to do on an early quiz is...

he likes to throw an If and Only IF (IFF) to see if we know that 2 proofs need to be done

Prove by CONTRAPOSITION: for all (upside-down A) N, if 3N+2 is odd, then N is odd.

if N is even, then 3n+2 is not odd, so 3n+2 is even -> then you prove using this new statement n=2k 3n=6k 3n+2 = 6k + 2 3n + 2 = 2 * (3k+1) proven indirectly using this claim using a much simpler proof, using contrapositive

in, for all (upside down A) positive integers A and B, if N = A*B, then A <= √(n) or B<= √(n), is the OR inclusive or exclusive

inclusive unless we say otherwise

to disprove a general statement

need to prove a counter-example (there exists)

to disprove a specific statement

need to supply a general proof that no such value exists (upside down a "for all", the specific statement doesn't hold)

DeMorgan's Law

negation of "n either has property X or property Y (or both)" is "n has neither property X nor Y"

synonym for assumptions

premises

Prove by Contradiction: There are arbitrarily large prime numbers

proof by contradiction: - assume there is a large prime number p - p! is divisible by all primes <=p (so, all primes by our assumption) - p! + 1 is divisible by no primes <=p - by Fundamental Theorem of Arithmetic, p!+1 must have a prime factorization, but each prime must be >p

to prove a specific statement (such as backwards E ["there exists"] an even prime), what do you need to do

provide an example

prove a general statement (for-all statement)

supply a general proof

Why is the statement "if 3n+2 is even, then N is even"

this is a different statement overall called the converse, not the logical opposite

opposite of "if n is a odd integer, then n^2 is also odd"

this statement isn't saying anything about even numbers, so you can't say if n is even logical statement of an if then statement will be more specific logical opposite is: n is odd AND n^2 is even NOT AN IF THEN STATEMENT

for all symbol

upside down A

definition of rational

where both numerator and denominator are integers

Why is Proof by Contradiction more powerful than Contraposition?

you can use it on problems that aren't just if then statements contrapositions only work for if then statements