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Proof troubles

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Let a be a positive number. Then there exists exactly one natural number b such that b++ =a.

    3. My attempt;

    Axiom 2.4
    Different natural numbers must have different successors; if n, m are natural numbers and n is not equal to m, then n++ is not equal to m++. equivalently if n++ = m++ then n = m.

    Let a be a positive number. Then there exists exactly one natural number such that b++ = a.


    PROOF

    Let P(b) be the statement Let a be a positive number. Then there exists exactly one natural number such that b++ = a..
    Then P(0) is true since 0++ = 1 > 0

    Suppose that c++ = d is true where c in N and d in N - {0}.

    Now we wish to deduce that (c++)++ = d++ is true. This is the case since (c++)++ is the increment a natural number and d++ is the increment of the positive number d. Hence the inductive is true and by Axiom 2.4 b is unique. Hence P(b) is true.

    Is this true? sloppy? false? total rubbish?

    I'm an undergrad who is new to analysis and proof.
     
    Last edited: Nov 13, 2014
  2. jcsd
  3. Nov 13, 2014 #2
    Do you have a well-defined concept for ">" at this stage? What is your definition of "positive number"?

    P, your alleged statement about b, is actually a statement about a.

    You might be better off with a contradiction proof, but there are two cases to dispose of: there is no such b, or there is more than one choice for b.
     
  4. Nov 13, 2014 #3
    no, your right, i don't!. I am guessing you know what textbook i am reading.
    It suggests that i use induction, but i have failed.
    to use a contradiction proof;

    Suppose there is no natural number b such that b++ = a.
    But this would suggest that the set of natural numbers is a finite set.
    this is a contraction of definition 2.1.1

    suppose that there exists more than one natural numbers such that b++ = a .
    that is b++=a and b++=c. where a is not equal to c.
    it follows that (b++)=a++ and (b++)=c++
    Now by axiom 2.4 a++=c++ implies that a = c, this is a contradiction.
    Hence there exists exactly one natural number b such that b++ =a.

    I feel that disposed both cases.


    Concerning my attempt at a proof by induction, when you say

    could you add a little more detail please.
     
  5. Nov 13, 2014 #4
    No, I don't have your textbook or any idea what it is :-)

    The whole proof is to establish that each non-zero natural number has a unique predecessor. So when we say:

    "Let a be a positive number. Then there exists exactly one natural number b such that b++ =a "

    ... that's describing a property of a - for any a that is a positive number.

    If your textbook suggests induction, you should probably go along with that. I don't really follow your contradiction proof, perhaps because of not knowing your textbook..
     
  6. Nov 13, 2014 #5
    If i was to tell you that definition 2.1.1 states that the set of natural numbers is an infinite set and axiom 2.4 is the one in the original post would that help? or is it just a terrible proof? haha. I honestly thought it proved that a unique natural number b exists such that b++=a

    in the proof by induction department...


    We shall induct on a.

    First we will do the base case where a = 1.
    Now 0++ = 1, this is the base case done.

    Now suppose inductively that b++ = a, now we must show that (b++)++ = a++ to close the induction.

    [since (b++)++ = a++ then by axiom 2.4 and the inductive hypothesis b++ = a. ]
    This seems like its all going well, up until the statement in the [ ].
    I don't see any other way to prove the inductive.
    Other than saying that a++ = a + 1. and (b++)++ = (b++) + 1.
    Then using the cancellation law and by the inductive hypothesis b++ = a.

    Then there is the task of proving that b is unique.
    but i thought i achieved that in the second part of my attempt at proof by contradiction.
     
  7. Nov 14, 2014 #6
    I meant to ask again what the definition of positive number is that you are using.

    If you were showing that b is unique, you would have something like:

    Assume b++ = a and c++ = a
    {some proof}
    Therefore b = c
    Therefore, for given a, there is at most one b such that b++ = a
     
    Last edited: Nov 14, 2014
  8. Nov 14, 2014 #7
    This is the definition i am using:

    a natural number n is said to be positive iff it is not equal to 0.

    i'm going to assume that my proof by induction is... ok haha.

    Thanks for the help! the book i am using is analysis 1 by Terence Tao.
     
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