Hello everyone,

Question

Trying to get my head around deductions and the deductive step using my text book, could someone look over my work:

Question:
n+6 is odd if and only if 5n+1 is even.

So my working, here it goes:

n+6=2k+1
n=2k-5

thus

5n+1=5(2k-5)+1
=2(5k-12)

so 5n+1 is even since 5k-12 is an integer.

5n+1=2k
subtract 4n and add +5 to both sides leaves.
n+6=2k-4n+5
=2(k-2n)+5

so n+2 is odd since k-2n is an integer.

Any of this making sense to anyone?

Thanks!

Answer 1

n+6 is odd if and only if 5n+1 is even

if 5n+1 is even, 5n is odd. So, n is odd (only odd*odd = odd)
If n is odd, n+6 is odd, (only odd+even=odd)

if n+6 is odd, n is odd, since odd numbers differ by 2.
so, 5n is odd, making 5n+1 even.

USING

odd*odd = odd since
(2m+1)(2n+1) = 2(2mn+m+n)+1 = 2k+1

even+odd = odd since
2m + 2n+1 = 2(m+n)+1 = 2k+1

Answer 2

I assume you are allowed to use the basic rules of addition and multiplication of odds and evens

odd + odd ---> even
odd + even --->odd
even + even ---> even

so if 5n + 1 is even: given
then 5n is odd: odd + 1 ---> even

odd*odd ---> odd
odd*even ---> even
even*even ---> even

so if 5n is odd, n has to be odd, since only odd*odd yields an odd

if n is odd, then n+1 is even, since
odd + odd---> even

Answer 3

my last 2 line should say:
if n is odd, then n+6 is odd, since
odd + even ---> odd