proof by contradiction

prove by contradiction that if n^2 is a multiple of 3 then n is a multiple of 3.

Assumption: There exists a number n such that n^2 is a multiple of 3, but n is not a multiple of 3. All multiples of 3 can be written in the form n = 3k where k is an integer, therefore 3k + 1 and 3k + 2 are not multiples of 3. Let n = 3k + 1 n^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1 In this case n^2 is not a multiple of 3. Let n = 3k + 2 n^2 = (3k + 2)^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k + 1) + 1 In this case n^2 is also not a multiple of 3. This contradicts the assumption that n^2 is a multiple of 3. Therefore if n^2 is a multiple of 3, n is a multiple of 3.

prove by contradiction that if p+q = odd then at least one of p and q is odd

Assumption: if p + q is odd than neither p nor q is odd. p is even, p = 2k q is even, q = 2m p + q = 2k + 2m = 2(k + m) So p + q is even. This contradicts the assumption that p + q is odd. Therefore, if p + q is odd then at least one of p and q is odd.

prove by contradiction that if pq is even then at least one of p and q is even

Assumption: if pq is even then neither p nor q is even. p is odd, p = 2k + 1 q is odd, q = 2m + 1 pq = (2k +1)(2m + 1) = 2km + 2k +2m +1 = 2(km + k + m) +1 So pq is odd. This contradicts the assumption that pq is even. Therefore, if pq is even then at least one of p and q is even.

prove by contradiction that there exists no integers a and b for which 21a +14b =1

Assumption: there exist integers a and b such that 21a + 14b = 1. Since 21 and 14 are multiples of 7, divide both sides by 7. So now 3a + 2b = 1/7 3a is also an integer. 2b is also an integer. The sum of two integers will always be an integer, so 3a + 2b is an integer. This contradicts the statement that 3a + 2b = 1/7 Therefore there exist no integers a and b for which 21a + 14b = 1

prove by contradiction that if n^3 is even then n is even

Assumption: there exists a number n such that n^3 is even but n is odd. n is odd so write n = 2k + 1 n^3 = (2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 + 6k^2 + 3k) +1 So n^3 is odd. This contradicts the assumption that n^3 is even. Therefore, if n^3 is even then n must be even.

prove by contradiction that if ab is an irrational number then at least one of a and b is an irrational number

a Assumption: if ab is an irrational number then neither a nor b is irrational. a is rational, a = c/d where c and d are integers. b is rational, b = e/f where e and f are integers. ab = ce/df, ce is an integer, df is an integer. Therefore ab is a rational number. This contradicts assumption that ab is irrational. Therefore, if ab is an irrational number then at least one of a and b is an irrational number.

prove by contradiction that if n^2 is odd then n is odd

negation: There exists a number n such that n^2 is odd but n is even n is even so write n = 2k n^2 = (2k)^2 = 4k^2 = 2(2k^2) So n^2 is even. This contradicts the assumption that n^2 is odd. Therefore, if n^2 is odd then n must be odd.

prove by contradiction that there are infinite number of prime numbers

negation: there are a finite number of prime numbers list of all prime numbers: p1, p2, p3, p4....., pn-1, pn N = p1 x p2 x p3...xpn +1 __________________________________ p3 remainder of 1 So N = either a prime or a prime factor not in the list of all possible prime numbers This is a contradiction to the negation Therefore there is an infinite number of prime numbers

prove by contradiction that there is no greatest even integer

negation: there is a greatest even integer, 2n 2(n + 1) is also an integer and 2(n + 1) > 2n 2n + 2 = even + even = even So there exists an even integer greater that 2n. This contradicts the assumption that the greatest even integer is 2n. Therefore there is no greatest even integer.

prove by contradiction that √2 is an irrational

negation: √2 is a rational number √2 = a/b for integers a and b; there are no common factors with a and b 2 = a^2/b^2 therefore a^2 = even and a = even if a = even = 2n, n = integer a^2 = 2b^2 therefore b^2 = even, b = even if a and b are even, they have a common factor of 2. this contradicts the statement that a and b have no common factors therefore √2 is an irrational