Is this discrete math question correct? Prove the following is true. If 7n+3 is odd, then n is even. *This is a proof by contrapositive?

Question

My answer: Suppose n is odd.

So n= 2k+1 for some integer k.

7n+3=7(2k+1)+3= 14k+10 = 2(7k+5)

By definition, 7n+3 is even.

Since the contrapositive is true, that implies the original statement is also true.

Is this correct?

Answer 1

Favourite answer

Yes.

The contrapositive of "if 7n + 3 is odd, then n is even" is "if n is odd, then 7n + 3 is even."

You have proven the contrapositive to be true, therefore the original statement is true.

{Another way to prove the statement without using contrapositives is:

(An odd number) minus (an odd number) equals (an even number).

If 7n + 3 is odd, then (7n + 3) - 3 = an even number.

Thus, 7n + 3 - 3 = 7n is an even numer.

(An odd number) times (an odd number) = (an odd number).

Therefore, if 7n is even, then n cannot be odd and must be even.}

Answer 2

The only thing missing is that it must also state that n is an integer in the original statement. Otherwise the statement isn't true.

The contrapositive of "If 7n+3 is odd, then n is even" would be "if n is not even then 7n+3 is not odd"

But what if n = 1/7? n is not even but 7n+3 is odd.

Answer 3

Yep, your steps are all correct for a contrapositive proof.

Answer 4

(7n + 3) is odd → then n is even ?

(7n + 3) ← if it's odd, you can divide it by 2

= (7n + 3)/2

= (7n/2) + (3/2)

= (7n/2) + [(2 + 1)/2]

= (7n/2) + (2/2) + (1/2)

= (7n/2) + 1 + (1/2) → if n is even, you can divide it by 2

= [7 * (n/2)] + 1 + (1/2) → if n is even, you can divide it by 2 → so (n/2) is an integer

= [7 * integer] + 1 + (1/2) → you can say that [7 * integer] is too an integer

= integer + 1 + (1/2) → and you can see that: integer + 1 is too an integer

= integer + (1/2) → but you know that 1 is not divisible by 2

As [integer + (1/2)] is not divisible by 2, so can say that: [integer + (1/2)] is odd

→ then: (7n + 3) is odd