1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving a theorum

  1. Jun 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Here's an excerpt from a book called "Algebra made easy". Needless to say I have taken algebra before and done well in it this is only the first chapter in the book, but I feel very lost at understanding this, its confusing.

    Here's another example of it. I just feel awful because I used to be very good at understanding algebra, in class with a teacher but learning it by myself again is very hard for me.

    Since multiplication is just a short cut to addition, I assume the rules are the same for multiplication

    Rules for adding odd and even numbers:
    If you add together two even numbers, the result is an even number
    If you add together two odd numbers, the result is an even number
    If you add together one even number and one odd number, the result is an odd number

    Here's what's confusing me:
    "Next, we developed our first result that we proved, rather than assumed. (a proved result is called a theorem.) We set out to prove the addition properties for odd and even numbers. We realized that any even number could be written in the form 2 x n, where n is some natural number. (I think a natural number times a natural number always equals a natural number. if you divide or subtract a natural number it doesn't)

    An odd number can be written in the form 2 x n + 1
    We tried adding together one even number called 2 x n and another even number called 2 x m, calling the result sum s

    s = (2 x m) + (2 x n)
    Using the distributive property,
    s= 2 x ( m + n)

    From the closure property, m + n must be a natural number, so s can be written in the form s = 2 x (some natural number)
    Therefore, s must be even
    Next, we tried adding together two odd numbers, called 2 x m + 1 and 2 x n + 1 (again calling the result s)
    s= (2 x m + 1) + (2 x n + 1)
    s = 2 x m + 2 x n + 2
    s = 2 x (m + n + 1)

    Since m + n + 1 must be a natural number, it follows that s must be even. We had one more combination to do: the sum of an odd number (which we called 2 x m +1 ) and an even number (which we called 2 x n):
    s = (2 x n ) + (2 x m + 1 )
    s = 2 x (m + n) + 1

    Since 2 x (m + n ) must be even, it follows that 2 x (m + n) + 1 must be odd.
    "We did it!" The professor exclaimed in amazement. "We can prove general behavior rules by using symbols to stand for letters! I wasn't even sure that could be done!"

    Can someone please try and explain this? I've always liked learning not only how to do things, but WHY you do them, and this explains it, yet I feel its very confusing for an introductury chapter on algebra.
  2. jcsd
  3. Jun 1, 2009 #2
    where n is a natural number

    i think one of the things that's confusing you is that they use "2*n + 1" in their definition of odd numbers and then go use "n+m" in the proof, so when you read through the equations in the proof and then go back to that definition of an odd number, you see "n" twice and each incarnation of it does not have to have the same value. The important thing that the definition in the quote is getting across is that "2 * natural number + 1" is an odd number, not that "2*n + 1" is an odd number. I think it might help a little to rewrite the definition of an odd number using a different letter. It doesn't affect anything but makes it a bit clearer to follow the proof, not needing to deal with two copies of "n" that mean different things.

    Rewrite: An odd number can be written in the form 2*a + 1 where "a" is a natural number

    Then go through the proof again
    s = (2*n) + (2*m + 1)
    s = (2*n + 2*m) + 1
    s = 2*(n+m) + 1
    Because "a" represents any natural number, and "n+m" is one specific natural number, you can just say "Suppose that a = n + m for the time being. It's legal for 'a' to be any natural number, and 'n + m' is a natural number, so it's legal to suppose that 'a' happens to be 'n + m' for the sake of our proof. Then whenever you see 'n + m' in your equations, you can substitute it out and plug in an 'a' in its place, for they both have the same value anyway."
    s = 2*a + 1
    'tis an odd number

    'hope that helped a bit. Have a nice day, sire.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Proving a theorum
  1. Binomial theorum (Replies: 8)

  2. De Moivre's Theorum (Replies: 6)

  3. Pythagoras theorum (Replies: 4)