Download Proof Solutions: Inclass worksheet

Proof Solutions: Inclass worksheet
1) Theorem: Let n ∈ Z, If n is even then n2 is even.
Proof: Given that n is even, by definition we know that n = 2a for a ∈ Z.
We will now consider n2 = n*n = (2a)(2a) = 2(2a2) = 2d where
integer d = 2a2.
Therefore by definition of even we have shown that n2 is even.
2) Given integers a, b, and c, If a bc → a b
This hypothesis is not valid as the following counter example demonstrates:
7 21 where 21 = bc and b = 3 and c = 7
3) Theorem: Given integers a, b, c, If a b and a (b + c) , Then a c
Proof: We are given that b = aq and b + c = ad by definition of divisible.
Since b + c = ad then c = ad – b = ad – aq = a(d – q) = ap where p is an
integer and p = d – q.
Conclusion: Since c = ap, then by definition of divisible we have shown that a c
4) Theorem: The product of two rational numbers is rational.
a
c
Proof: Suppose that n = , m = are two rational numbers where a,b,c,d are
b
d
a c ac
integers and b ≠ 0 and d ≠ 0. Now consider n*m = * =
b d bd
Notice that ac = k and bd = h are both product of integers and therefore are each integers.
k
The product m*n = satisfies the definition of a rational number. So the product of two
h
rational numbers is rational.
5) Typo Corrected: Theorem: Given an integer n, If n is even that n2 is divisible by 4.
Proof: If n is even then n = 2a for some integer a (by definition of even)
Consider n2= n*n = (2a)(2a) = 4a2 = 4d where d = a2 and d is an integer.
Therefore n2 is divisible by 4 by definition of divisible.
6) Theorem: If n is a natural number then n2 + n is even.
Proof: Consider n2 + n = n(n + 1), since n and n + 1 are consecutive
integers, one must be even and the other odd. We have already
proved that the product of an even integer and any other integer is
even. Therefore n2 + n is even.
7) Theorem: Given n is an integer, n is even if and only if 7n + 4 is even.
Case I: If n is even, then 7n + 4 is even:
Proof: If n is even then n = 2a for some integer a.
So 7n + 4 = 7(2a) + 4 = 2[7a + 2] = 2m for m = 7a + 2 and
m is an integer. There by definition 7n + 4 is even
Case II: If 7n + 4 is even, then n is even.
Proof: (First attempt)
Since 7n + 4 is even then 7n + 4 = 2d for some integer d.
2d − 4
Now solve for n to get n =
(unfortunately it is not easy to
7
verify that this is an even integer. So we need to start again.)
(Second Attempt) Proof by Contrapositive:
Proof: If n is odd, then 7n + 4 is odd
If n is odd, we have that n = 2a + 1.
Now consider 7n + 4 = 7(2a + 1) + 4 = 14a + 7 + 4 = 14a + 10 +1
= 2(7a + 5) + 1 = 2k + 1 which is odd by definition.
Since we have shown the contrapositive is true. The original
statement is equivalent to the contrapositive and also true.
8)
Prove using “ proof by contradiction”:
9)
Proof similar to 6)
10) Theorem: If a, b, c are prime numbers greater than 2, then a 3 + b3 ≠ c3
Proof: Since a, b, c are prime and greater than 2 then they are each odd.
Therefore a 3 , b3 , c3 are each odd since the product of odd integers
are also odd.
The sum a 3 + b3 = 2a for some integer a, since the sum of two
odd integers is even. Therefore a 3 + b3 ≠ c3 since c3 has already
been shown to be odd.
11) Theorem:
If x is a real number, x = 1 if and only if x3 − 3x 2 + 4 x − 2 = 0
Part I: If x = 1 and x is a real number, then x3 − 3x 2 + 4 x − 2 = 0
Proof: Evaluate at x = 1 to get (1)3 − 3(1) 2 + 4(1) − 2 = 0
Part II: If x3 − 3x 2 + 4 x − 2 = 0 , then x = 1is a real solution.
Proof: Factor x3 − 3x 2 + 4 x − 2 = ( x − 1)( x 2 − 2 x + 2) = 0
The solutions are x = 1 and x = (1 ± i) and x = 1 is the real solution
Therefore we have proved our conjecture.
12) Invalid step is found when you have division by (a – b) which is really division by 0