Results 1 to 5 of 5

Math Help - Do not quite understand this simple proof

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    48

    Do not quite understand this simple proof

    Claim: There exists a number \alpha such that \alpha ^2 = 2

    Proof:
    So, we look at the set S = \{x \in R | x^2 < 2\}. We want to consider \alpha = sup(S). Claim that \alpha ^2 = 2, and prove by contradiction.

    There are two possibilites: \alpha ^2 < 2 or \alpha ^2 > 2.

    For \alpha ^2 < 2: Let n \in R be large enough so that n > \frac{2\alpha + 1}{2 - \alpha ^2} (<---Do not understand choice of n here). Then \frac{1}{n} < \frac{2 - \alpha^2}{2\alpha + 1}. Then

    (\alpha + \frac{1}{n})^2 = \alpha^2 + \frac{2\alpha}{n} + \frac{1}{n^2} < \alpha^2 + \frac{2\alpha}{n} + \frac{1}{n} = \alpha^2 + \frac{2\alpha + 1}{n}

    So \alpha^2 + \frac{2\alpha}{n} < \alpha^2 + (2\alpha + 1) \frac{2 - \alpha^2}{2\alpha + 1} = 2

    But then \alpha + \frac{1}{n} \in S, and \alpha + \frac{1}{n} > \alpha, contradicting the fact that \alpha is an upper bound for S.

    This isn't the complete proof. You then have to use P.B.C. for the other case, \alpha > 2. But, my problem is that I don't understand the choice of n here. Of course it makes sense looking at the proof, but how does one know how to choose n for any other problem, say, for \alpha^3 = 4?
    Last edited by Ares_D1; August 4th 2010 at 10:54 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,311
    Thanks
    1286
    I rather suspect that this choice of n was done by "trial and error"! (Many proofs are created by trial and error, then written out with showing the "errors".) Of course, it is not really the "n" that is important- it is \frac{1}{n}= \frac{2n+1}{2- \alpha^2} that we want. We want to prove that if \alpha^2< 2, then \alpha is not an upper bound for the set given. 2- \alpha^2 is the amount by which \alpha^2 is less than 2 and the denominator was chosen so that (\alpha+ \frac{2-\alpha^2}{2\alpha+1})^2 is still less than 2.

    You say "it makes sense looking at the proof" and I suspect that they just tried 2- \alpha^2 divided by something to get a fraction of the interval and tried different divisors until they got something that worked.

    Here's how I would do that, with a slightly different "estimate".

    Suppose \alpha^2 is less than 2. Then we must have \alpha^2= 2- h for some h. Use that h to find some number larger than \alpha whose square is still less than 2.

    First note that 1.4^2= 1.96< 2 and 1.5^2= 2.25> 2 so we can assume that 1.4< \alpha< 1.5 and that h< 2- 1.96= 0.04.

    What about the number half way between? (\alpha+ h/2)^2= \alpha^2+ 2(h/2)\alpha+ h^2= \alpha^2+ h\alpha+ h^2= \alpha^2+ h(\alpha+ h). No, that won't work- \alpha+ h> 1.4> 1, (\alpha+ h)h> h so (\alpha+ h)^2> \alpha^2+ h= 2.

    What about \alpha+ h/3? (\alpha+ h/3)^2= \alpha^2+ 2(h/3)\alpha+ h^2/9= \alpha^2+ h(\frac{2}{3}\alpha+ h/9). \frac{2}{3}\alpha> \frac{2}{3}1.4= .933 and \frac{2}{3}\alpha< \frac{2}{3}(1.5)= 1.0. Since we have to add h^2/9, that still isn't good enough- we might have \frac{2}{3}\alpha+ h^2/9 larger than 1 so that (\alpha+ \frac{h}{3})^2= \alpha^2+ h(\frac{2}{3}\alpha+ \frac{h^2}{9})> \alpha+ h= 2.

    Okay, what about \alpha+ h/4? (\alpha+ h/4)^2= \alpha^2+ \frac{1}{2}h\alpha+ \frac{h^2}{16} = \alpha^2+ h(\frac{\alpha}{2}+ \frac{h}{16}.

    Since 1.4< \alpha< 1.5, .7< \alpha< .75 and since h< .04, \frac{h}{16}<0.0025, \frac{\alpha}{2}+ \frac{h}{16}< 0.7525< 1 so \alpha^2+ h(\frac{\alpha}{2}+ \frac{h}{16})< \alpha^2+ h= 2. Yes, we have a number, larger than \alpha whose square is still less than 2 contradicting the fact that \alpha is the least upper bound of that set.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    16
    People usually find these proofs in different way and then when they put them in this form, they make less sense.

    The important parts of this proof are that you are choosing \alpha = \sup(S). You know that \alpha^2 should equal 2, but you want to assume on the contrary. Once you do that, you want to pick a very small number, say \epsilon and add it to \alpha. You want to show that this is also in the set S and thus this contradicts your choice of \alpha.

    Given \alpha^2<2[/Math], you want [Math](\alpha + \epsilon)^2 = \alpha^2+2\alpha\epsilon +\epsilon^2 < 2. Since \epsilon is supposedly very small, it is definitely < 1, so \epsilon^2 < \epsilon. So you can sub this to get

    \alpha^2+2\alpha\epsilon +\epsilon^2 < \alpha^2 + (2\alpha+1)\epsilon.

    You want to make this <2 so you might as well try to get it to look like \alpha^2 +(2 - \alpha^2). comparing the two expressions, you want to make (2\alpha+1)\epsilon < (2 - \alpha^2).

    Well, just divide out to get \epsilon < \frac{2-\alpha^2}{2\alpha+1}. when you sub this in for epsilon, you will get what you want. All this shows that you didn't have to choose and n and stuff, which makes it somewhat better to understand I think.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    The choice of n looks in first instance a bit arbitrary, but it's chosen such that the little trick works...

    We can pull the same trick for all c> 0 and showing the existence of \alpha s.t \alpha^3= c

    We'll decide what we want n to be afterwards...

    Let S = \left\{x\in R: x^3< c\right\} and \alpha  = \sup(S). And let's assume \alpha^3< c.

    First observe that \alpha\geq 0 and ...

    (\alpha+\frac{1}{n})^3< \alpha^3+\frac{1}{n}(3\alpha^2+3\alpha+1)

    So that if we want our little trick to work we need \frac{1}{n}< \frac{c-\alpha^3}{3\alpha^2+3\alpha+1}

    And since \alpha \geq 0 , this number is positive and it's obvious we can find such n.

    Now we have (\alpha+\frac{1}{n})^3< \alpha^3+\frac{1}{n}(3\alpha^2+3\alpha+1) < \alpha^3+\frac{c-\alpha^3}{3\alpha^2+3\alpha+1}(3\alpha^2+3\alpha+1  )=c

    since \alpha+\frac{1}{n}> \alpha this contradicts the fact that \alpha= \sup(S)

    We may do something likewise in the case \alpha^3> c
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    16
    More general case. Let \alpha \in \mathbb{R}, you want to show [LaTeX ERROR: Convert failed] for some real and positive c and let it be >1.

    You let the set be [LaTeX ERROR: Convert failed] . Let \alpha = \sup(S).

    Then we want  (\alpha+\epsilon)^n < c.<br />
    We need a little result before that. We have \epsilon<1. Then We have \epsilon^2<\epsilon. So by induction you will have that \epsilon^k<\epsilon for all  k \in \mathbb{N}.

    This leads to

    \displaystyle{(\alpha+\epsilon)^n = \alpha^n + \sum_{k=1}^n \binom{n}{k}\alpha^{n-k}\epsilon^k < \alpha^n+\sum_{k=1}^n \binom{n}{k}\alpha^{n-k}\epsilon = \alpha^n+\epsilon\sum_{k=1}^n \binom{n}{k}\alpha^{n-k} = \alpha^n +\epsilon D}

    where D is the sum. You want \epsilon D < c - \alpha^n, so you choose

    \displaystyle{\epsilon < \frac{c-\alpha^n}{D}}
    Last edited by Vlasev; August 5th 2010 at 06:28 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: September 24th 2010, 05:47 AM
  2. Can you help me understand this logic proof?
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: February 24th 2010, 05:51 AM
  3. Don't understand the proof of a simple theorem
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 15th 2009, 12:55 PM
  4. Replies: 1
    Last Post: December 13th 2008, 03:27 AM
  5. Can't understand proof.
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: December 7th 2008, 08:02 AM

Search Tags


/mathhelpforum @mathhelpforum