# Math Help - Integers

1. ## Integers

Hello,

I am trying to solve this problem:

If j and k are integers and j - k is even, which of the following must be even?

a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j

According to the booklet the answer is d, however how would I even go about arriving at this answer? Any help would be appreciated...

2. ## Re: Integers

Originally Posted by fsiwaju
Hello,

I am trying to solve this problem:

If j and k are integers and j - k is even, which of the following must be even?

a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j

According to the booklet the answer is d, however how would I even go about arriving at this answer? Any help would be appreciated...
1. Set $j = k + 2m, m\in \mathbb{N}$. Then (j - k) is even.

2. Check all 5 terms, replacing j by k + 2m. For instance:

$j + 2k = k + 2m + 2k = 3k + 2m~\implies~$ ..... j+2k is only even if k is even because then j is even too.

...

$jk + j = (k+2m)k + (k + 2m) = (k + 2m)(k + 1)~\implies~$..... (jk + j) is even if k is even because then the 1st bracket would be even too; (jk + j) is even if k is odd because then the 2nd bracket would be even too. A product is even if one factor is even.

3. ## Re: Integers

If j and k are integers and j - k is even, which of the following must be even?

a) k
b) jk
c) j + 2k
d) jk + j
e) jk - 2j
if j - k is even, then j and k are both even or both odd.

4. ## Re: Integers

if j - k is even either j and k are both odd, or j and k are both even.

if j and k are both odd, j + 2k is odd, so the answer cannot be (c). similar considerations rule out (a) and (b).

(e) is a little trickier, but only a little: if j and k are both odd, so is jk, in which case jk - 2j will be odd, too.

so (d) is all that's left. but let's not be lazy. let's PROVE jk + j MUST be even.

case (1):

j and k are both even.

then jk is even, and j is even, so jk + j is even.

case (2):

j and k are both odd.

then jk is odd, and j is odd, so jk + j is even.

*************

to fully appreciate this, you might want to know why j - k is even implies BOTH have to be even or odd.

to prove THAT, we need to prove "the arithmetic of parity" (even/oddness):

even+even = even
even+odd = odd
odd+odd = even

even*even = even
even*odd = even
odd*odd = odd.

well, this is a chore, but we only have to prove it ONCE, and then we can use it for the rest of our lives.

even + even = ?

we can write an even number as 2m, so let's call our two even numbers 2m and 2n.

2m + 2n = 2(m+n) <---the 2 in front makes this an even number.

even + odd = ?

2m + (2n+1) = (2m + 2n) + 1 = 2(m+n) + 1, taking t = m+n, we see we have a number of the form 2t+1, which is odd.

odd + odd = ?

(2m+1) + (2n+1) = 2m + (1+2n) + 1 = 2m + (2n+1) + 1 = (2m + 2n) + (1+1) = 2(m+n) + 2 = 2(m+n+1) <---this is even.

even*even = ?

(2m)(2n) = 4mn = 2(2mn) <--even

even*odd = ?

(2m)(2n+1) = 4mn + 2m = 2(2mn+m) <--even

odd*odd = ?

(2m+1)(2n+1) = (2m)(2n+1) + (1)(2n+1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1 <--odd.

we can ALMOST prove j-k even implies j,k both even, or both odd. we need one more step:

if t is even, so is -t:

if t is even t = 2n, so -t = (-1)(t) = (-1)(2n) = -2n = 2(-n) <---even.

if t is odd, so is -t:

if t is odd, t = 2n+1, so -t = (-1)(t) = (-1)(2n+1) = -2n - 1 = -2n + (-1) = -2n + (1 - 2) = -2n + (1 + -2) = -2n + (-2 + 1) = (-2n + -2) + 1 = 2(-n - 1) + 1<-- odd.

so now we look at the 4 possible cases for j and k:

j odd, k even:

then -k is even, so:

j - k = j + -k = odd + even = odd.

j even, k odd:

then -k is odd, so j - k = j + -k = -k + j = odd + even = odd.

j even, k even:

then j - k = j + -k = even + even = even <---half-done

j odd, k odd:

then j - k = j + -k = odd + odd = even <---all done.

Thanks !!!!