1. Limited time only! Sign up for a free 30min personal 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!

Proof by contradiction for simple inequality

  1. Jan 10, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm trying to show that if ##a \approx 1##, then
    $$-1 \leq \frac{1-a}{a} \leq 1$$

    I've started off trying a contradiction, i.e. suppose
    $$ \frac{|1-a|}{a} > 1$$

    either i)
    $$\frac{1-a}{a} < -1$$
    then multiply by a and add a to show
    $$1 < 0$$
    which is clearly false,

    or ii)
    $$1 < \frac{1-a}{a}$$
    Which by the same reckoning leads me to
    $$a < \frac{1}{2}$$

    Which seems inconsistent with the original statement that ##a \approx 1##.
    It's hardly a rock solid proof though, is it?
    Is there a better way of doing this?
    Thanks
     
  2. jcsd
  3. Jan 10, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, it is inconsistent! This is a proof by contradiction, remember? Your result that a< 1/2 contradicts the fact that a is close to 1.
     
  4. Jan 10, 2014 #3
    Thanks HallsofIvy. So you think that is enough to prove the original statement then?
    The reason I thought it was a bit wooly, was that ##a \approx 1## is unclear regarding exactly how close to 1 it is!
    Also, given that my final conclusion limits ##a## to less than 1/2, I'm unclear as to why they chose 1 as the bound in the original question. They could have chosen a range of other numbers (i.e ##-0.5 < (1-a)/a < 0.5##) and normally these things work out neater.
    But if you think it's a sound proof, that's all that counts.
    Thanks
     
  5. Jan 10, 2014 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, your conclusion is NOT that a is less than 1/2! Your conclusion is that assuming that (1- a)/a> 1 leads to "a< 1/2" which contradicts the given fact that "[itex]a\approx. 1[/itex]".
     
  6. Jan 10, 2014 #5
    Sorry, I didn't make myself clear:
    Yes, I am concluding that assuming (1-a)/a > 1 leads to a < 1/2. But my problem is with the second point. Does that necessarily contradict the fact that ##a \approx 1##? That doesn't seem mathematically rigorous to me (although I'm only a beginner). Imagine, for example, that ##a## was a measure of the distance on a galactic scale...something like a=1/4 would be fairly close to 1, on that scale wouldn't it? Doesn't the "approximately equal to" have some more precise definition?
    Thanks
     
  7. Jan 10, 2014 #6

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    You can make it more precise by defining "If ##a \approx 1## then X" to mean "There is an ##\epsilon > 0## such that X for all ##a## satisfying ##|a - 1| < \epsilon##."

    Then ##a < 1/2## contradicts ##a \approx 1## because for any candidate ##\epsilon##, you can find a value ##a## such that ##1/2 < a < 1## for which ##|a - 1| < \epsilon## but not ##a < 1/2##.
     
  8. Jan 10, 2014 #7
    Thanks CompuChip, that tidies it up nicely.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Proof by contradiction for simple inequality
  1. Proof by contradiction (Replies: 2)

  2. Proof by contradiction (Replies: 6)

  3. Proof by contradiction (Replies: 4)

  4. Proof by contradiction (Replies: 5)

Loading...