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
Hello everyone,
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!
3 answers
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
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
my last 2 line should say:
if n is odd, then n+6 is odd, since
odd + even ---> odd
if n is odd, then n+6 is odd, since
odd + even ---> odd