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**Proof Must Have Statement of what is to be proven.**

"Proof:" to indicate where the proof starts Clear indication of flow Clear indication of reason for each step Careful notation, completeness and order Clear indication of the conclusion

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**Number Theory - Ch 3 Definitions**

Z --- integers Q - rational numbers (quotients of integers) rQ a,bZ, (r = a/b) ^ (b 0) Irrational = not rational R --- real numbers superscript of positive portion only superscript of negative portion only other superscripts: Zeven, Zodd , Q>5 "closure" of these sets for an operation Z closed under what operations?

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**Integer Definitions even integer odd integer prime integer (Z>1)**

n Zeven k Z, n = 2k odd integer n Zodd k Z, n = 2k+1 prime integer (Z>1) n Zprime r,sZ+, (n=r*s) (r=1)v(s=1) composite integer (Z>1) n Zcomposite r,sZ+, n=r*s ^(r1)^(s1)

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**Constructive Proof of Existence**

If we want to prove: nZeven, p,q, r,sZprime n = p+q ^ n = r+s ^pr^ ps^ qr^ qs let n=10 n Zeven by definition of even Let p = 5 and the q = 5 p,q Zprime by definition of prime 10 = 5+5 Let r = 3 and s = 7 r,s Zprime by definition of prime 10 = 3+7 and all of the inequalities hold Make sure to mention the appendix that has rules of algebra.

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**Methods of Proving Universally Quantified Statements**

Method of Exhaustion prove for each and every member of the domain rZ+ where 23<r<29 p,q Z+ (r = p*q)^(p<=q) Generalizing from the "generic particular" suppose x is a particular but arbitrarily chosen element of the domain show that x satisfies the property i.e. rZ, r Zeven r2 Zeven Exhaustion: Let r be an arbitrary member of Z 24 = 4* Assume r Zeven 25 = 5*5 exists k in evens such that r = 2k (def of even) 26 = 2* when squaring both sides of r = 2k, we get r2 = 4k2 27 = 3*9 by factoring out a 2 we get r2 = 2(2k2) 28=2* k2 is even because 2k2 is an integer by closure therefore r2 is even r in Zeven -> r2 in Zeven

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**Examples of Generalizing from the "Generic Particular"**

The product of any two odd integers is also odd. m,n Z, [(m Zodd n Zodd ) m*n Zodd ] The product of any two rationals is also rational. m,nQ, m*n Q

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**Disproof by Counter Example**

rZ, r2Z+ rZ+ Counter Example: r2= 9 ^ r = -3 r2Z+ since 9 Z+ so the antecedent is true but rZ+ since -3Z+ so the consequent is false this means the implication is false for r = -3 so this is a valid counter example When a counter example is given you must always justify that it is a valid counter example by showing the algebra (or other interpretation needed) to support your claim

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**Division definitions d | n kZ, n = d*k n is divisible by d**

n is a multiple of d d is a divisor of n d divides n standard factored form n = p1e1 * p2e2 * p3e3 * …* pkek

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**Proof using the Contrapositive**

For all positive integers, if n does not divide a number to which d is a factor, then n can not divide d.

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**Proof using the Contrapositive**

For all positive integers, if n does not divide a number to which d is a factor, then n can not divide d. n,d,cZ+, ndc nd

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**Proof using the Contrapositive**

For all positive integers, if n does not divide a number to which d is a factor, then n can not divide d. n,d,cZ+, ndc nd n,d,cZ+, n|d n|dc proof: let n, d and c be arbitrary in Z+ assume n|d there exists a k in Z such that d=nk by def of divides multiplying both sides by c gives dc = nkc kc in Z by closure of Z in multiplication n|dc by definition of divides n|d -> n|dc by closing conditional world w/out contra forall n,d,c in Z+, n|d -> n|dc by generalizing from GP

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**more integer definitions**

div and mod operators n div d --- integer quotient for n mod d --- integer remainder for (n div d = q) ^ (n mod d = r) n = d*q+r where n Z0, d Z+, r Z, q Z, 0 r<d relating “mod” to “divides” d|n 0 = n mod d 0 d n definition of equivalence in a mod x d y d|(x-y) [note: their remainders are equal] sometimes written as x y mod d meaning (x y) mod d examples of equivalence in a modulous 122,12 mod 10 & 10,13 mod 3

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**Quotient Remainder Theorem**

nZ dZ+ q,rZ (n=dq+r) ^ (0 r < d) Proving definition of equiv in a mod by using the quotient remainder theorem This means prove that if [m d n], then [d|(n-m)] where m,nZ and dZ+ Using what you know about mod operator from programming that the remainder is the same exists k,l,r in Z such that n = dk+r and m = dl+r n=dk+r n-dk = r m = dl + r m-dl = r n-dk = m-dl n-m = dk-dl n-m = d(k-l) k-l in Z by closure so d| (n-m) by definition of divides

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**Proofs using this definition**

mZ+ a,bZ a m b k Z a=b+km mZ+ a,b,c,dZ a m b ^ c m d a+c m b+d # #2 m in Z+ and a,b in Z are GP m in Z+, a,b,c,d in Z are GP Assume a m-equiv b assume a m-equiv b and c m-equiv d m|(a-b) by def of equiv in a mod m | (a-b) and m|(c-d) by def of equiv in a mod exists k in Z such that (a-b) = mk exists k in Z such that a-b = km a = b + km by algebra exists x in Z such that c-d = xm by def of divides Assume exists k in Z a=b+mk a-b = mk (a-b)+(c-d) = km+xm by adding equals to equals Since k is an integer, (a+c)-(b+d) = (k+x)m by algebra m|(a-b) by def of divides Since k+x in Z by closure of Z in addition a m-equiv b by def of equiv in a mod m| (a+c)-(b+d) by definition of divides (a+c) m-equiv (b+d) by def of equiv in a mod

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**Proof by Division into Cases**

nZ 3n n2 3 1 # #2 forall n in Z [n/2] is Let n be arbitrary in Z n/2 if n is even We know by the quotient remainder theorem that or (n-1/2) if n is odd there exists a q and r such that proof: n = 3q+0 or n = 3q+1 or n=3q+2 Case 1:Assume n is even exists k in Z n = 2k Taking these three cases: by def of even Case 1: (assume: n=3q+0) exists x in Z where x = floor n/2 n=3q and 3|n which means the antecedent is false x <= n/2 < x and the implication is true x <= 2k/2 < x+1 x <= k < x+1 x = k Case 2: (assume n = 3q+1) Case 2: Assume n is odd since n=3q+1, n^2 (3q+1)^2=9q^2+6q+1=3(3q^2+2q)+1 exists w in Z n = 2w n^2-1 = 3(3q^2+2q) by def of odd Since 2q^2+2q is in Z by closure of Z in * and + exists x in Z when x = floor n/ | n^2 –1 by the definition of divides x <= n/2<x and n^2 3-equiv to 1 by def of equiv in a mod x <=(2w+1)/2<x+1 x<= w+1/2 < x Case 3: (assume n = 3q+2) x <= w < x+1 since n=3q+2,n^2=(3q+2)^2=9q^2+18q+4=3(3q^2+6q+1)+1 x = w n^2 – 1 = 3(3q^2+6q+1) n = 2w q^2+6q+1 is in Z by closure of Z in * and + n-1 = 2w 3|n^2-1 by definition of divides (n-1)/2 = w n^2 equiv to 1 by def of equiv in a mod (n-1)/2 = x implication true in all three cases

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**Floor and Ceiling Definitions**

n is the floor of x where xR ^ n Z x = n n x < n+1 n is the ceiling of x where xR ^ n Z x = n n-1 < x n

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**Floor/Ceiling Proofs x,yR x+y = x + y xR yZ x+y = x +y**

#1 disproof by counter example #2 proof by GP let x = 2.5 and let y = let x be arbitrary in R and let y be arbitrary in Z then floor(x) = 2 and floor(y) =2 by the definition of floor of x and floor(x)+floor(y) = exists n in Z such that n <= x<n+1 adding y to all sides of the inequality we get x+y = n+y <= x+y < n+y+1 then floor(x+y) = by closure n+y and n+1+y are in Z so this is the definition of the floor of x+y which are not equal

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**Proof by Division into Cases (again)**

The floor of (n/2) is either a) n/2 when n is even or b) (n-1)/2 when n is odd # #2 forall n in Z [n/2] is Let n be arbitrary in Z n/2 if n is even We know by the quotient remainder theorem that or (n-1/2) if n is odd there exists a q and r such that proof: n = 3q+0 or n = 3q+1 or n=3q+2 Case 1:Assume n is even exists k in Z n = 2k Taking these three cases: by def of even Case 1: (assume: n=3q+0) exists x in Z where x = floor n/2 n=3q and 3|n which means the antecedent is false x <= n/2 < x and the implication is true x <= 2k/2 < x+1 x <= k < x+1 x = k Case 2: (assume n = 3q+1) Case 2: Assume n is odd since n=3q+1, n^2 (3q+1)^2=9q^2+6q+1=3(3q^2+2q)+1 exists w in Z n = 2w n^2-1 = 3(3q^2+2q) by def of odd Since 2q^2+2q is in Z by closure of Z in * and + exists x in Z when x = floor n/ | n^2 –1 by the definition of divides x <= n/2<x and n^2 3-equiv to 1 by def of equiv in a mod x <=(2w+1)/2<x+1 x<= w+1/2 < x Case 3: (assume n = 3q+2) x <= w < x+1 since n=3q+2,n^2=(3q+2)^2=9q^2+18q+4=3(3q^2+6q+1)+1 x = w n^2 – 1 = 3(3q^2+6q+1) n = 2w q^2+6q+1 is in Z by closure of Z in * and + n-1 = 2w 3|n^2-1 by definition of divides (n-1)/2 = w n^2 equiv to 1 by def of equiv in a mod (n-1)/2 = x implication true in all three cases

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**Prime Factored Form kZ, p1,p2,…pkZprime, e1,e2,…ekZ+,**

n = p1e1 * p2e2 * p3e3 * …* pkek Unique Factorization Theorem (Theorem 3.3.3) given any integer n>1 kZ, p1,p2,…pkZprime, e1,e2,…ekZ+, where the p’s are distinct and any other expression of n is identical to this except maybe in the order of the factors Standard Factored Form pi < pi+1 mZ,8*7*6*5*4*3*2*m=17*16*15*14*13*12*11*10 Does 17|m ??

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**Steps Toward Proving the Unique Factorization Theorem**

Every integer greater than or equal to 2 has at least one prime that divides it For all integers greater than 1, if a|b, then a (b+1) There are an infinite number of primes #1 Proof by div into cases #2 let a and be be arbitrary in Z where a > 1 case 1: n is prime assume a|b it has a prime that divides it exists k in Z such that b = ak (itself) assume a | b+1 case 2: n is not prime exists m in Z such that b+1=am we need chapter 4 to prove this ak+1 = am by substitution 1 = am-ak=a(m-k) 1/a = m-k 1/a is an integer only when a = 1 but m-k is an integer by closure Contradiction a not| b+1 #3 assume there are a finite number of primes lets say the number of primes is m We can take that list of primes and find their product p1*p2*p3*…*pm = k k is an integer by closure of integers in multiplication Lets look at k+1 since p1|k, p1 not| k+1 since p2|k, p2 not|k+1 … but by the first proof on this page there must be an integer that divides k+1 that isn’t in this list of m integers contradiction so there must be a finite number of primes

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**Using the Unique Factorization Theorem**

Prove that the Prove: aZ+qZprime q|a2 q |a #1 Assume the cuberoot(3) is rational there exists a,b in Z such that cuberoot(3) = a/b and b neq 0 cubroot(3) = a/b b(cuberoot(3)) = a 3*b^2 = a^2 by the prime factorization theorem a can be prime factorized to p1^e1*p2^e2*…px^ex b can be prime factorized to q1^d1*q2^d2*…qy^dy a^2 = p1^2e1*p2^2e2*…px^2ex b^2 = q1^2d1*q2^2d2*…qy^2dy By the unique prime factorization theorem one of the p’s must be 3 factor all of the three’s out a^2 = 3^2ei(a1^2) where a1 is the left over product of primes and it is important to know that 3 not| a1 3*b^2=3^2ei(a1^2) cancel a 3 from both sides and we get b^2=3^(2ei-1)(a1^2) Since ei >=1, there is still a 3 left on the right so one of the q’s must be a 3 factor all of the three’s out to get b^2 = 3^2dj(b1^2) where b1 is the left over product of primes and it is important to know that 3 not|b1 by substitution we get 3^2dj(b1^2) = 3^(2ei-1)(a1^2) by the unique prime factorization theorem 2dj = 2ei-1 contradiction since one is odd the other is even Prove: a,b,cZ+pZprime [(a2 = b3 = c) ^ (p|c)] (p6 | c) (a^2=b^3=c) and (p|c) since p|c – exists k in Z such that c=pk Since a^2 = c a can be prime factorized and a^2 has 2 p’s since b^3 = c and b can be prime factorized b^3 has 3 p’s since c must have a multiple of 2 p’s and a multiple of 3 p’s it must have a multiple of 6 p’s

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**Summary of Proof Methods**

Constructive Proof of Existence Proof by Exhaustion Proof by Generalizing from the Generic Particular Proof by Contraposition Proof by Contradiction Proof by Division into Cases

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**Errors in Proofs Arguing from example for universal proof.**

Misuse of Variables Jumping to the Conclusion (missing steps) Begging the Question Using "if" about something that is known

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