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All x for which 2^x < 8

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Find all x for which 2x < 8 (where x is a natural number).


    2. Relevant equations

    Those pesky 12 properties.


    3. The attempt at a solution

    This could be done by simply testing values of x, but that approach is not very scalable. What is the correct way to start this (using only the basic properties of numbers)? What is Spivak looking for? Thanks.
     
  2. jcsd
  3. Feb 4, 2012 #2
    It's easy to find all x for which this holds by trial-and-error. Then prove that any other (larger) values of x give > instead of <.
     
  4. Feb 4, 2012 #3
    As I said, I was wondering if there was another way. Trial and error doesn't seem very 'proofy' to me.
     
  5. Feb 4, 2012 #4

    Dick

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    I don't know the pesky 12 properties. But try writing it as 2^x<2^3. Any pesky properties to help with that?
     
  6. Feb 4, 2012 #5
    12 Properties. But really, any number that makes you happy. I'm not really sure what to do with this. There is no basic property that says that i can bring down the exponents. This problem doesn't really matter in the long run, I was just curious as to what Spivak expects me to do with it.
     
  7. Feb 4, 2012 #6
    I think Dick's hint is spot on. I think his approach is what Spivak expected. Can you see what to do next?
     
  8. Feb 4, 2012 #7
    Dick's hint seems good, but if you can show by exhaustion that some x > some other number doesn't work, well, exhaustion is infact a valid proof. See, four color theorem.
     
  9. Feb 4, 2012 #8

    Office_Shredder

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    Prove that the function 2x is an increasing function on the natural numbers directly: if f(x)=2x, show that f(x)<f(x+1) for all x. Then it's clear that 2x<23 if and only if x<3
     
  10. Feb 4, 2012 #9
    Hello folks :smile:

    @Office_Shredder: Is that induction? That is next chapter, so I wanted to hold off on that.

    @Joriss: So when there is a contradiction, exhaustion is acceptable proof, but if there is no contradiction, hen it is not. Is that a true statement>

    @DivisionByZro: Unfortunately, no. the only thing I can think to do with that is to write out what 2^x and 2^3 means:

    2x < 23

    2*2 x-number-of-times < 2*2 3-times

    is that what you mean?
     
  11. Feb 4, 2012 #10
    No, exhaustion works for contradictions or when there are finite number of possibilities. I didn't say exhaustion is a good route here, just that when applicable it is a legitimate proof (which, if you already knew, my bad :)).
     
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