Math 143
Weds 10/13/10
Sec 20 & 21 Board Work Assignments
1. (Colby -- 10/14) Prove by method of smallest counterexample: For all natural numbers
n, 6 | n 3  n .
2. (Adam -- 10/14) Prove by induction: Let x be a real number, x ≠ 1. Then for all natural
numbers n with n ≥ 1, 1  x  x    x
2
n 1
xn 1
.

x 1
3. (Dung -- 10/14) Prove by method of smallest counterexample: For all natural numbers n
n(n  1)( 2n  1)
with n ≥ 1, 12  2 2  3 2    n 2 
.
6
4. (Scott -- 10/14) Prove by induction: For all natural numbers n with n ≥ 1, n!  n n .
5. (Moana -- 10/14) Prove by method of smallest counterexample: For all natural numbers
n with n ≥ 4, 2 n  n ! .
6. (Darrin -- 10/15) Prove by induction: For all natural numbers n with n ≥ 4, n 2  2 n .
7. (CJ -- 10/15) Prove by method of smallest counterexample: For all natural numbers n
with n ≥ 1, n!  n n .
8. (Tina -- 10/15) Prove by induction: For all natural numbers n with n ≥ 1,
9 *1  9 *10  9 *100    9 *10 n 1  10 n  1 .
9. (Aaron -- 10/15) Prove by method of smallest counterexample: For all natural numbers
n, n  2 n .
10. (Nathan -- 10/15) Prove by induction: For all natural numbers n with n ≥ 4, 2 n  n ! .
Outline: Proof by method of smallest counterexample.
Theorem: For all n in S, P(n). [S must be well-ordered.]
Proof:
 Assume the theorem is false, so there is at least one counterexample in S.
 Let n be the smallest counterexample in S.
 min(S), the smallest element in S, is NOT a counterexample because it makes
P(n) true [justify]; so n > min(S).
 Consider n-1, or the largest element of S which is less than n. Since n is the
smallest counterexample, n-1 is NOT a counter example. Use the fact that P(n-1)
is true to show P(n) is true.
 Contradiction! P(n) is false, since n is a counterexample, but also P(n) is true as
shown above. Therefore, the statement has no counterexamples in S. The
theorem is true.
Outline: Proof by induction.
Theorem: For all n in N with n  n0 , P(n).
Proof:
 First check the base case: P(n0) is true because ... [justify]
 Let n in N be arbitrary ( n  n0 ).



Assume P(n) is true.
Work forward with valid deductions to show that P(n+1) is true.
Thus P(n) implies P(n+1). By the principle of mathematical induction, P(n) is
true for all n with n  n0 .