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Math Help - Integers

  1. #1
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    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...
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  2. #2
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    Re: Integers

    Quote Originally Posted by fsiwaju View Post
    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.
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  3. #3
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    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.
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  4. #4
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    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 from earboth
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  5. #5
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    Re: Integers

    Thanks !!!!
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