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Doing proofs using the basic properties of numbers, problems

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that if 0 < a < b,
    then a < √ab < (a + b)/2 < b


    2. Relevant equations
    To prove this use the 12 properties of numbers (commutativity, trichotomy law, associativity, etc...).


    3. The attempt at a solution
    The main problem is I don't know if I need to define the square root function since I'm doing a proof. I'm new to all this, this is my first time doing analysis, so bear with me.
    I assume you break up the inequality into parts, since a<b is already defined, we don't need to prove the whole inequality. So I did the first part, unintutively

    a<√ab
    a^2 < (ab)
    (a)(a) (a^-1) < (ab)(a^-1)
    a<b which was assumed thus the inequality is proven

    Also
    (a + b)/2 < b
    (a +b)(2^-1) (2) < b(2)
    a < b

    Problem is I don't know what to write division of 2 as, since it's not listed in basic property. Also I don't know how to prove the 2 centre terms of the inequality: √ab < (a + b)/2
     
  2. jcsd
  3. Sep 18, 2012 #2

    micromass

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    This is a Spivak problem, am I right? Your concerns are completely justified, you need to define square roots and division (and even the number 2) before you can do this problem. Spivak doesn't do this (he does it in a later chapter though). I think Spivak wants this to be a challenging exercise using the math intuition you already have, rather than an exercise in proving things from the 12 axioms. He does this kind of thing quite a lot in the first chapters of the book. For example, in chapter 1 he uses things like [itex]4^x[/itex] in the exercises, which is also not defined. In short: you got to take things on faith in the first few chapters of the book, they will be rigorized later.

    Anyway, let me define the terms for you:
    • If [itex]c\geq 0[/itex], then we define [itex]\sqrt{c}[/itex] as the unique positive number whose square in c. So, it is a number such that [itex]\sqrt{c}\geq 0[/itex] and [itex](\sqrt{c})^2=c[/itex]. Again, you need to take on faith that such a number actually exists and that is not unique. This is serious because it cannot follow from the first 12 axioms (indeed, [itex]\sqrt{2}[/itex] does not exist in [itex]\mathbb{Q}[/itex], but [itex]\mathbb{Q}[/itex] satisfies all axioms). We need another axiom to ensure existence of square roots. Spivak introduces this axiom later in his book. Just take on faith now that the square root exists.
    • The number 2 is just defined as 1+1. There is no problem here.
    • Division is defined as multiplying with the inverse. So [itex]a/b[/itex] is defined as [itex]ab^{-1}[/itex] (as long as b is nonzero). So [itex](a+b)/2[/itex] is defined as [itex](a+b)*2^{-1}[/itex]. There is no problem here since it follow from the twelve axioms that every nonzero number has a unique inverse. We do need to check that 2 is nonzero.

    I hope this helps.
     
  4. Sep 18, 2012 #3
    Yes thankyou that clarifies alot! And ya it is a Spivak problem. But I'm still confused for the second part where I have to prove √ab < (a + b)/2, or am I supposed to figure it out myself ;). Basically what I mean is am I allowed to square both sides, since i didn't read on about square roots in the book. I know I'm an idiot. :shy:
     
  5. Sep 18, 2012 #4

    Dick

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    If, as micromass says, you don't have to worry about axioms, then sure, just square both sides and use algebra. I don't know that you are an idiot. You'll have to prove it :).
     
  6. Sep 18, 2012 #5

    micromass

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    Yeah, you should square both sides. You should be able to prove why [itex]0\leq a\leq b[/itex] implies [itex]a^2\leq b^2[/itex] using the 12 axioms.

    The only thing you need to accept is [itex]\sqrt{ab}\geq 0[/itex] and [itex](\sqrt{ab})^2=ab[/itex] (and that such number actually exists and is unique).
     
  7. Sep 19, 2012 #6
    okay thanks for the help, greatly appreciated.
     
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