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Section 1.6: #17
Show that if n is an integer and n^3 + 5 is odd, then n is even using:
a) Contraposition
Remember: the contraposition of p ⇒ q is ¬q ⇒ ¬p and if you know one to be
true then you also know the other is true.
Also Remember: The opposite of odd is even and vice versa so “not odd” is the
same as saying even (so long as you’re talking about numbers).
In this case “n^3 + 5 is odd” is p and “n is even” is q. So the contrapositive
version of the statement would be: “if n is odd then n^3 + 5 is even.”
So we have “n is odd” as premise.
Odd numbers can always be represented as “2k + 1” for some integer k.
Therefore n = 2k + 1 for some integer k.
Then we can substitute 2k + 1 for n in “n^3 + 1”
n 3 + 5 = (2k + 1) 3 + 5
We then have:
n 3 + 5 = (8k 3 + 12k 2 + 6k + 1) + 5
n 3 + 5 = 8k 3 + 12k 2 + 6k + 6
n 3 + 5 = 2(4k 3 + 6k 2 + 3k + 3)
At this point you can set some other variable equal to the mess inside the
parentheses but it is not necessary. As long as the number is conclusively reduced to 2
times some integer (which the mess insides the parentheses is since it is only addition and
multiplication of known integers. Be careful of division though), you can claim the
number is even So with n being odd we have reduced n^3 + 5 to an even number and
therefore proven the contradiction.
b) Contradiction
Remember: for contradiction, you take what you are trying to prove and suppose
the opposite is true. You then go through some logical steps to arrive at a conclusion that
is contradictory to the premise or some basic rule of mathematics.
So in this case we are trying to prove that n is even. So we suppose the opposite is
true:
Suppose n is odd.
Then n = 2k + 1 for some integer k.
Then substitute 2k + 1 in for n in “n^3 + 5.”
n 3 + 5 = (2k + 1) 3 + 5
We then have:
n 3 + 5 = (8k 3 + 12k 2 + 6k + 1) + 5
n 3 + 5 = 8k 3 + 12k 2 + 6k + 6
n 3 + 5 = 2(4k 3 + 6k 2 + 3k + 3)
And just like last time, we have concluded that n 3 + 5 is even. This is a
contradiction with the premise. Therefore n cannot be odd. Thus it must be even.
Section 1.6: #20 (altered to ask about P(n) instead of P(1) )
Prove proposition P(n) where P(n) is the proposition “If n is a positive integer, then
n 2 ≥ n .” What kind of proof did you use?
Go for proof by contradiction here. You do so by assuming the opposite of the
conclusion you’re looking for and using logical steps to find a contradiction with the
premise of the question or a rule of mathematics. So…
Suppose n 2 < n
(Opposite of greater than/equal is less than)
Divide both sides by n to get:
n <1
Since n is an integer, the greatest possible for n in this case is 0.
This is a contradiction with the premise that states n is a positive integer.
Therefore n 2 cannot be less than n .
Therefore: n 2 ≥ n
Section 1.7: #4
Use a proof by cases to show that min(a, min(b, c)) = min(min(a, b), c) whenever a,
b, and c are real numbers.
The question asks for a proof by cases so the first step is to look at this program
and identify the possible cases.
In this question, the cases can be divided up based whether a is greater than b and
whether b is greater than c and so on. This leads to six cases:
a≥b≥c
a≥c≥b
b≥c≥a
b≥a≥c
c≥b≥a
c≥a≥b
So the next step is prove the statement for every case:
For a ≥ b ≥ c :
min(a, min(b, c)) = min(min(a, b), c)
min(a, c) = min(b, c)
c=c
For a ≥ c ≥ b :
min(a, min(b, c)) = min(min(a, b), c)
min(a, b) = min(b, c)
b=b
For b ≥ c ≥ a :
min(a, min(b, c)) = min(min(a, b), c)
min(a, c) = min(a, c)
a=a
For b ≥ a ≥ c :
min(a, min(b, c)) = min(min(a, b), c)
min(a, c) = min(a, c)
c=c
For c ≥ b ≥ a :
min(a, min(b, c)) = min(min(a, b), c)
min(a, b) = min(a, c)
a=a
For c ≥ a ≥ b :
min(a, min(b, c)) = min(min(a, b), c)
min(a, b) = min(b, c)
b=b
So since all possible cases turn out to be true, the statement is true. The key to a
proof by cases is, naturally, picking a group cases both covers all possibilities given the
premises of the question and conveniently divides the problem up into parts that can be
solved.