Number theory

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Number theory Method of proofs

• Isaac Fung

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Summary

• Direct proof
• Perfect square
• Divisibility
• Quotient Remainder Theorem
• Rational numbers
• Proof by contradiction
• Proof by cases
• Proof by contrapositive

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Direct proof 3.1.42

• If integers m and n are perfect squares, then
  is also a perfect square
• Proof
• Since m and n are perfect squares, ma2 and nb2
  for some integers a and b.
• So is also a perfect
  square

A perfect square is the product of some integer
and itself
Dont forget to assume a and b to be integers
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Direct proof 3.1.38

• Any product of 4 consecutive integers is one less
  than a perfect square
• 1st attempt Consider k, k1, k2, k3 and try to
  find n s.t.
• n2 1 (k)(k1)(k2)(k3)
• 2nd attempt Consider k-1, k, k1, k2
• (k-1)(k)(k1)(k2)1 k42k3-k-2k1
• If this equals (ak2-bkc)2 for some integers a,
  b and c,
• then we are done
• Trial and error shows that a 1, b 1, c -1.

This is just a sketch but not a proof!
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Direct proof 3.1.38

• Proof
• (k-1)(k)(k1)(k2) k42k3-k2-2k1-1
• (k2k-1)2-1
• Since k is an integer, k2k-1 is also an
  integer.
• So, product of any 4 consecutive integers is one
  less than some perfect square
• Another proof
• (k-1)(k2)k2k-2, (k1)(k) k2k
• Let m k2k-1, so (k-1)(k2) m-1, (k1)(k)
  m1
• (k-1)(k)(k1)(k2) (m-1)(m1) m2-1
• So, product of any 4 consecutive integers is one
  less than some perfect square

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Disproof by counterexample 3.1.44

• If p is a prime number, then 2p 1 is also a
  prime number
• This is false as 211 1 2048 1 23(89)

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Direct proof 3.2.25 (Supplementary)

• Suppose a, b and c are integers and x, y, z are
  non-zero real numbers that satisfy the following
  equations
• and and
• Then x is rational
• Proof
• By the first 2 equations, we have
• Substituting them into the last equation gives

Note x /b, x/c, otherwise y, z are not real
numbers
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Direct proof 3.2.25 (Supplementary)

• By assumption, x/0, so
  . Therefore x is rational.

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Direct proof 3.2.28 (Supplementary)

• If z is a root of a polynomial with rational
  coefficients, then z is a root of a polynomial
  with integer coefficients
• Proof
• Suppose z is a root of a polynomial with rational
  coefficients.
• Then where d is the degree
  of the polynomial,
• ai and bi for i0 to d are integers

An polynomial in a variable x is a sum of powers
of x. e.g. c3x3c2x2c1xc0. c3, c2, c1 and c0
are called the coefficients of the polynomial. A
real number z is its root if c3z3c2z2c1zc00
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Direct proof 3.2.28 (Supplementary)

• (continued)
• Let and ci aiB/bi for i 0 to d
• Note ci is an integer for i 0 to d
• So is a polynomial with integer
  coefficients
• Moreover
• So z is also a root of this polynomial

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Direct proof 3.3.40

• Suppose n nknk-1n1n0 is a decimal number.
• If the sum of digits is divisible by 9,
  then n is divisible by 9
• Observation
• Pick an arbitrary multiple of 9, e.g. 1476
• Observe that 1476 10001 1004 107 16
• (9991)1 (991)4 (91)7
  16
• 9(111111417) 1 4 7 6

This is the sum of digits of n
This is a multiple of 9
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Direct proof 3.3.40

• Suppose n nknk-1n1n0 is a decimal number.
• If the sum of digits is divisible by 9,
  then n is divisible by 9
• Proof

where m0 0, m1 1, m2 11, m3 111
Therefore 9n iff sum of its digits is divisible
by 9
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Direct proof 3.3.42

• Suppose n nknk-1n1n0 is a decimal number.
• If the sum of digits is divisible by 11,
  then n is divisible by 11
• Observation
• Pick an arbitrary multiple of 11, e.g. 1485
• Observe that 1485 10001 1004 108 15
• (1001-1)1 (991)4 (11-1)8
  15
• 11(9119418) - 1 4 - 8 5

This is the difference between the sum of odd
digits and the sum of even digits of n
This is a multiple of 11
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Direct proof 3.3.42

• Suppose n nknk-1n1n0 is a decimal number.
• If the sum of digits is divisible by 11,
  then n is divisible by 11
• Proof

where m00, m11, m29, m391, m4909, m59091
Therefore 11n iff sum of its digits is divisible
by 11
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Direct proof 3.4.33

• If m, n and d are integers and d(m-n), then m
  mod d n mod d
• Proof
• Suppose m , n and d are integers s.t. d(m-n)
• Let r1 m mod d, so m k1d r1 for some
  integer k1
• Let r2 n mod d, so n k2d r2 for some
  integer k2
• Since d(m-n), m - n k3d for some integer k3
• So k1d r1 k2d - r2 k3d
• r1 - r2 (k3 k1 k2)d
• But by definition of remainder, 0
• So r1- r2

By the Quotient-Remainder Theorem
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Direct proof 3.4.34

• If m, n, a, b and d are integers and m mod d a
  and n mod d b, then (mn) mod d (ab) mod d
• Proof
• Since m mod d a , m k1d a for some integer
  k1
• Similarly, n k2d b for some integer k2
• Let r (ab) mod d, so a b k3d r for some
  integer k3
• Note that 0 r
• m n k1d a k2d b
• k1d k2d k3d r
• (k1 k2 k3) d r
• By the Quotient-Remainder Theorem, the unique r
  s.t. mnk4dr and
• 0 r(ab) mod d

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Proof by contradiction 3.6.22

• If a and b are rational numbers, b/0, and r is
  an irrational number, then abr is irrational
• Proof
• Assume to the contrary that a and b are rational,
  b/0 and r is an irrational number and abr is
  rational, i.e. abrc1/c2 for some integers c1
  and c2.
• abr c1/c2
• br c1/c2 a d1/d2 for some integers d1 and
  d2, since difference between 2 rational numbers
  is rational
• But r d1/(d2b) is rational because a rational
  divided by a rational is rational. This
  contradicts the assumption that r is irrational
  and completes the proof.

Why direct proof does not work?
This is the negation of the given statement
we can divide it by b as b/0 by assumption
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Proof by contradiction 3.7.8

• If n is any integer that is not a perfect square,
  then is irrational
• Proof
• Suppose for the sake of contradiction that n is
  an integer that is not a perfect square and
  is rational, that is for some
  coprime integers a and b.
• Then
• But a and b are coprime, so a2 and b2 are also
  coprime. Also n is an integer, so b1 and n a2.
  However this contradicts the assumption that n is
  not a perfect square

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Proof by contrapositive 3.7.8

• If is rational, then n is a perfect square
• Proof
• Suppose is rational, then for
  some coprime integers a and b.
• Then
• But a and b are coprime, so a2 and b2 are also
  coprime. Also n is an integer, so b1 and n a2.
  Therefore n is a perfect square

This is the contrapositive of the last theorem
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Proof by contradiction 3.7.9

• log2(3) is irrational
• Proof
• Assume to the contrary that log2(3) m/n for
  some integers m and n.
• However 3n is a power of 3, so it has only factor
  of 3. But by the Fundamental Theorem of
  Arithmetic, this is the unique factorization of
  3n. Therefore 3n cannot equal 2m. This is a
  contradiction.

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Proof by cases 3.7.19

• Prove that if p1, p2, , and pn are distinct
  prime numbers with p12 and n 1, then
  p1p2pn1 can be written in the form of 4k3
  for some integer k
• Proof
• Let p p1p2pn1. Since p1 2, p is odd.
• Any odd integer can be written as either 4k1 or
  4k3 for some integer k.
• Consider the case p 4k1 for some integer k.
• Then p1p2pn 4k. But p1 2, so p2pn
  2k.
• However, p1, p2, , and pn are assumed to be
  distinct prime numbers. p2, p3 , and pn are all
  odd. There is a contradiction.
• So it must be the case that p can be written in
  the form of 4k3 for some integer k

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3.7.20

• For all integers n, if n 2, then there is a
  prime number p such that
• n
• Proof
• Let m n!-1. Recall that n! 123n, so
  every prime number n is a factor of n!.
• Also a prime factor of an integer k cannot be an
  factor of k-1 and k1. (the proof is similar to
  that of Proposition 3.7.3)
• So m has no prime factors n.
• Now if m itself is prime, then we are done, since
  n 2.
• In case m is not a prime, then m has a prime
  factor f n and f
• This is the prime we want.