Solution to 2. (i) Prove that if n is a positive integer, then n is even if … - Sikademy
Author Image

Archangel Macsika

2. (i) Prove that if n is a positive integer, then n is even if and only if 7n + 4 is even. (ii) Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd. (iii) Prove that m2 = n2 if and only if m = n or m = -n. (iv) Prove or disprove that if m and n are integers such that mn = 1, then either m = 1 or else m = -1 and n = -1.

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

2.(i)=> We prove first the direct implication. Assume n is even. Then n=2k for some integer k. Then 7n+4=7(2k)+4=14k+4=2(7k+2), which is even.


<= For the converse, which is “if 7n+4 is even, then n is even”, we use a proof by contrapositive.

The contrapositive is: “if n is not even (that is, odd), then 7n+4 is not even (that is, odd)”. If n is odd, then n=2k+1, for some integer k.

Then 7n+4=7(2k+1)+4=14k+7+4=14k+11, which is odd.

Hence we have proven that n is even if and only if 7n+4 is even. 


(ii) => We will use a direct proof on “If n is odd, then 5n+6 is odd”. Assume n is odd, so n=2k+1 for some integer k.

Then 5n+6=5(2k+1)+6=10k+5+6 =10k+11=2(5k+5)+1.

Thus, 5n+6 is odd.


<= We now must prove the converse, “If 5n+6 is odd, then n is odd.” For this, we will use proof by contrapositive. So the statement becomes “If n is not odd, then 5n+6 is not odd.” Assume that n is not odd, so n=2k for some integer k.

Then 5n+6=5(2k)+6=10k+6=2(5k+3).

Thus, 5n+6 is not odd.

Hence we have proven that n is odd if and only if 5n+6 is odd. 


(iii) => If m=n, then m^2=n^2 follows using basic principles of Algebra.

Similarly, if m=-n, then m^2=(-n)^2=n^2 follows using basic principles of Algebra.

<= Suppose that m^2=n^2. Notice that the square of any real number is non-negative, so we can take the square root of each side of this equation in order to obtain: \sqrt{m^2}=\sqrt{n^2}, or |m|=|n|. Thus, using basic principles of algebra, either m=n or m=-n.


(iv) mn=1=>m\not=0, n\not=0.

Case 1: Suppose |m|>1. Regardless of what n is, mn=1, so n=\dfrac{1}{m}. But we said that both m and n are integers. If |m|>1, then n here cannot be an integer. So, contradiction.

Case 2: Suppose |n|>1. Regardless of what m is, mn=1, so n=\dfrac{1}{m}. But we said that both m and n are integers. If |n|>1, then m here cannot be an integer. So, contradiction.

Case 3: Suppose m=1 and n=-1. Clearly then mn=-1. So, contradiction.

Case 4: Suppose m=-1 and n=1. Clearly then mn=-1. So, contradiction.

Case 5: Suppose m=1 and n=1. Then mn=1.

Case 6: Suppose m=-1 and n=-1. Then mn=1.

What techniques did we use? Cases and contradiction


Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-3889-qpid-2588