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Help Proof of simple theorems with addition and mulitplication axioms

  1. Feb 14, 2012 #1
    Thanks in advance. 1st day at calculus teacher wants proofs. They seem rudimentary but I've never done them and he doesn't help so I'm hoping someone here could please.

    These are the axioms:
    Addition:
    For a, b, and c taken from the real numbers
    A1: a+b is a real number also (closure)
    A2: There exist 0, such that 0 + a = a for all a (existence of zero - an identity)
    A3: For every a, there exist b (written -a), such that a + b = 0 (existence of an additive inverse)
    A4: (a + b) + c = a + (b + c) (associativity of addition)
    A5: a + b = b + a (commutativity of addition)
    Multiplication:
    For a, b, and c taken from the real numbers excluding zero
    M1: ab is a real number also (closure)
    M2: There exist an element, 1, such that 1a = a for all a (existence of one - an identity)
    M3: For every a there exists a b such that ab = 1
    M4: (ab)c = a(bc) (associativity of multiplication)
    M5: ab = ba (commutativity of multiplication)
    D1: a(b + c) = ab + ac (distributivity)


    Prove these theorems using the Addition and Mult. Axioms:
    Theorem #1: (-1) • a = -a
    Theorem #2: if a • b = 0 and a ≠ 0, then b = 0
    if a ≠ 0 and b ≠ 0, then a * b ≠ 0
    Theorem #3: if a ≠ 0 and b ≠ 0 then 1/a •*1/b = 1/a•b

    Here is the example he did in class:
    prove a•0=0
    line 1 a•0
    line 2:A2: a•0+0
    line 3:A3: a•0+{a•0+[-(a•0)]}
    line 4:A4: {a•0+a•0}+[-(a•0)]
    line 5:D1: a•(0+0)+[-(a•0)]
    line 6:A2: a•0+[-(a•0)]
    line 7:A3: 0

    I've never done proofs this way so if anyone has any pointers or ways of thinking about it that would really help. Professor is no help. Thanks guys/girls.
     
  2. jcsd
  3. Feb 14, 2012 #2

    SammyS

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    Hopefully by now, you know the rules of this Forum.

    Show us what you've tried, and where you're stuck, so we can help you.
     
  4. Feb 14, 2012 #3
    Ouch, not the answer I was hoping for. Ok I didn't want to overwhelm you with text since giving all the necessary information is long enough, but here's what I've tried:

    line 1:(-1)•a
    line 2:a•(-1)
    line 3:a•(-1)•(-1)•1
    line 4:(-1•1)•(a•1)
    line 5:-1•1•1•a+[a+(-a)]

    and then I go further down the rabbit hole of wtf, why is it so hard for me to prove -1•a=-a

    another attempt:
    (-1)•a
    (-a)
    -a
    yay! -a! but that's too easy, and I guess it's not using the axioms. Basically I'm lost and if you could just give me a place to start or a thing to keep in mind it would help. Telling me the rules of the forum is not helping. Trust me, I'm not trying to get anyone to do my homework, I do calc for fun. I just don't understand proofs yet. Please help!
     
  5. Feb 14, 2012 #4
    ok I think I'm on to something.. maybe? so lost so who knows.

    (-1)•a=-a
    line1: (-1)•a + -a + a axiom of additive inverse
    line2: -1•(a+a)+-a associative axiom
    line3: -1a+-1a+-a D1
    line4: -a axiom of additive inverse again?

    Is that right? Can I go from line 3 to line 4? -1a+-1a=0, right? Is there a step I'm missing?
     
  6. Feb 14, 2012 #5

    SammyS

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    First of all, you're not giving an Axiom number with each step.

    Line 3 doesn't follow from Line 2.

    Here's an idea to get you started down the right path:
    Can you show that (-1)•a+a is 0 ? ... using the axioms, of course.​
     
  7. Feb 14, 2012 #6
    Thanks SammyS. Sigh.. I can't.. When I look at that I just think to Distribute it.

    (-1)•a+a=0
    ...

    I just stare at it blankly. Can you give me another hint? Or do you know of a good reference I could read on this subject by any chance? I really appreciate your help. Thank you so much.
     
  8. Feb 15, 2012 #7

    SammyS

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    Sorry for the delay.

    You know that 1•a is the same thing as a, right?

    Starting with

    (-1)•a+a

    M2: (-1)•a+(1)•a

    Now factor out a. (That's D1 in the axioms.)

    Can you finish it?


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

    Now, to show that (-1)•a is the same thing as -a.

    Starting with (-1)•a . You can add 0.

    (-1)•a + 0 .[STRIKE] a + -a .Then[/STRIKE] Of course a + -a is 0 so:

    (-1)•a + (a + -a) , 1•a is the same as also use assoc law.

    ( (-1)•a + 1•a ) + -a . Now, factor out a, That's D1.

    Can you finish? You will need to use that 0•a is 0 , which isn't in an Axiom, but the proof is in your Original Post.

    Added in Edit:

    I see I had a typo above. I fixed it in red.
     
    Last edited: Feb 15, 2012
  9. Feb 15, 2012 #8
    Wow thanks. Picking up where you left off:

    M2: (-1)•a+(1)•a
    D1: a(-1+1)
    A3: a•0 using additive inverse. Now I can use Theorem 1 that he proved or write it out:
    A2: a•0+0
    A3: a•0+{a•0+[-(a•0)]}
    A4: (a•0+a•0)+[-(a•0)]
    D1: a•(0+0)+[-(a•0)]
    A2: a•0+[-(a•0)]
    A3: 0


    ******************
    D1 a•(-1•1) +-a
    Additive inverse: a•0 +-a
    Theorem 1: 0+-a
    Additive Inverse: -a


    Did I do it? Still shaky but feeling much better about it. Thanks a ton Sammy, bless your heart.
     
  10. Feb 15, 2012 #9

    Deveno

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    normally, i "work backwards" and when i finish, write my proof "upside-down":

    0 = 0 (obvious, right?)
    0 = a + (-a) (A3)
    0•a = a + (-a) (your "Theorem 1")
    (1 + (-1))•a = a + (-a) (using A3 applied to the real number 1)
    a•(1 + (-1)) = a + (-a) (M5, because D1 has the sum "on the right")
    a•1 + a•(-1) = a + (-a) (D1)
    1•a + (-1)•a = a + (-a) (M5 twice, on the left)

    now, there is a theorem needed to finish this...it's not a hard theorem, but it is really, really, important:

    IF:

    c + a = c + b

    THEN:

    a = b.

    well, you may say, this is obvious...we just "subtract c from both sides." there is one small problems with this:

    what does it mean to "subtract c" (subtraction is not defined in your axioms)?
     
  11. Feb 15, 2012 #10

    SammyS

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    I put a few comments in red above.

    But you did well.
     
  12. Feb 15, 2012 #11

    SammyS

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    This is all well and good. It's an excellent method for seeing your way through what steps are needed ...

    but
    ...

    At this point in OP's course, it's apparent that the professor wants the students to work with a string of equivalent expressions; i.e. there are no equations are to be used.
     
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