Download Monday, August 8: Samples of Proofs

Samples of Proofs
Direct Proof: (p  q) Prove that if n is odd, n2 is odd.
Let n be an odd number. Since n is odd, n = 2k + 1 where k I. N2 = (2k + 1)2 =
4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Since 2,k I and integers are closed with respect to
addition and multiplication, 2k2 + 2k I. So 2(2k2 + 2k) + 1 is an odd integer. Hence, if
n is an odd number, n2 is an odd number.
Indirect Proof: (q  p) Prove that if (x  a)(x  b)  0, then x  b.
Assume that x = b. Then x  b = 0. Furthermore, (x  a)(x  b) = 0. Hence the
contrapositive of our statement is true, so if (x  a)(x  b)  0, then x  b.
Proof by Contradiction: ( p  q)  ( r  r) Prove that if 3x + 5 = 14, then x  4.
Let 3x + 5 = 14 and assume x = 4. Then by substitution, 34 + 5 = 12 + 5 = 17, and
17  4. Hence, if 3x + 5 = 14, x  4.
Proof by Cases: (p  p  p p…)  q Prove that if n is an odd integer not divisible by
4, then n2 has a remainder of 1 when it is divided by 4.
Let n be an odd integer not divisible by 4. Then the remainder when n is divided by 4
must be 1 or 3. Let's consider the 2 cases that exist.
Case #1: (The remainder is 1.) Let n = 4k + 1, where k I. Then n2 = (4k + 1)2
2
= 16k + 8k + 1 = 4(4k2 + 2k) + 1. Since integers are closed with respect to addition and
multiplication and 4, 2, and k I, 4(4k2 + 2k) + 1 will have a remainder of 1 when it is
divided by 4.
Case #1: (The remainder is 3.) Let n = 4k + 3, where k I. Then n2 = (4k + 3)2
2
= 16k + 24k + 9 = 4(4k2 + 6k + 2) + 1. Since integers are closed with respect to addition
and multiplication and 4, 2, 6, and k I, 4(4k2 + 6k + 2) + 1 will have a remainder of 1
when it is divided by 4.
Existence Proof: ( Find a number that fits the conditions.) Prove that if a  b = a + b +
4, that there exists some number k so that a  k = k  a = a for all a  Reals.
Let a  b = a + b + 4. Then a  k = a + k + 4 and if a + k + 4 = a, then k = -4. Consider
k = 4. a  4 = a + 4 + 4 = a for all real values of a, and 4  a = 4 + a + 4 = a for
all real numbers a. Hence if a  b = a + b + 4 , then a  4 = 4  a = a for all
a  Reals.
Proof by Induction: (Set up the domino principle.) Prove that 3 + 7 + 11 + … + 4n  1
= 2n2 + n for all n  N.
Consider the statement: 3 + 7 + 11 + … + 4n  1 = 2n2 + n
(Prove that 1 works.) Since 3 = 212 + 1 = 3, this statement is true when n = 1.
(Set up the dominos.) Let the statement be true when n = k. Then,
3 + 7 + 11 + … + 4k  1 = 2k2 + k (Add the next term to both sides.)
+ 4(k+1)  1
+ 4(k+1)  1
3 + 7 + 11 + … + 4(k+1)  1 = 2k2 + k + 4(k+1)  1 = 2k2 + 5k + 3
= (2k2 + 4k + 2) + (k + 1) = 2(k2 + 2k + 1) + (k + 1)
= 2(k + 1)2 + (k + 1).
So when any number k works, k + 1 works. So by the principle of mathematical
induction, 3 + 7 + 11 + … + 4n  1 = 2n2 + n.