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    Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.
  
Root Beer. Topic: Real Analysis. November 22nd, 2006

Problem: (Stanford Math 51H) Prove that every positive real number has a positive square root (That is, for any b > 0 , prove that there is a real number \alpha such that \alpha^2 = b ). [Usual properties of the integers are assumed.]

Solution: Consider the set S = \{x \in \mathbb{R}: x > 0, x^2 < b\} . We can show that S is non-empty by selecting an integer n large enough such that \frac{1}{n^2} < b . Since b is a real and the integers are unbounded, there exists a positive integer k such that k^2 > b , thus x^2 < k^2 \Rightarrow x < k so S is bounded from above.

Now there must exist \alpha such that \sup S = \alpha . We claim that \alpha^2 = b . Suppose the contrary; then either \alpha^2 < b or \alpha^2 > b .

CASE 1: If \alpha^2 < b , then consider \left(\alpha+\frac{1}{n}\right)^2 = \alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2} . Choose n large enough such that

\alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2} < b

which is definitely true for any n > \frac{2\alpha+1}{b-\alpha^2} . But then \alpha+\frac{1}{n} \in S and \alpha+\frac{1}{n} > \alpha = \sup S , contradicting the fact that \alpha is the supremum.

CASE 2: If \alpha^2 > b , then consider \left(\alpha-\frac{1}{n}\right)^2 = \alpha^2-\frac{2\alpha}{n}+\frac{1}{n^2} . Again, choose n large enough such that

\alpha^2-\frac{2\alpha}{n}+\frac{1}{n^2} > b

which we know is true for n > \frac{2\alpha}{\alpha^2-b} (furthermore, we impose the restriction \alpha > \frac{1}{n} so our resulting real is positive). Then x \le \alpha-\frac{1}{n} for all x \in S and \alpha-\frac{1}{n} < \alpha = \sup S , contradicting the fact that \alpha is the supremum.

Hence we know that \alpha^2 = b , or that the square root of any positive real number exists and is a positive real number. QED.

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Comment: This is from the real analysis portion of a freshman honors calculus course, i.e. a rigorous treatment of the real numbers which is the basis for calculus like limits and stuff. Really understanding calculus involves really understanding how the real number system works.

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Practice Problem: (Stanford Math 51H) If a, b \in \mathbb{R} with a < b , prove:

(a) There is a rational r \in (a,b) .
(b) There is an irrational c \in (a,b) .
(c) (a,b) contains infinitely many rationals and infinitely many irrationals.

Posted in Real Analysis || 4 Comments »

4 Responses to “Root Beer. Topic: Real Analysis.”

  1. t0rajir0u Says:
    November 23rd, 2006 at 1:35 am

    how about this:

    the function f(x) = x^2 is continuous everywhere. since it takes the value f(0) = 0, which is less than b, and since it is unbounded, so it takes a value greater than b, then it takes on the value of b at some point in (0, infty)

  2. t0rajir0u Says:
    November 23rd, 2006 at 1:39 am

    er okay i guess proving that x^2 is continuous is kind of a stronger result isn’t it heh…

    a) let m be the smallest integer such that |ma – mb| > 1. then there exists some integer n such that ma < n < mb and hence a < n/m < b

    b) buhhh.

    c) obvious by induction. now if only i could prove b lol

  3. Xuan Says:
    November 23rd, 2006 at 3:52 pm

    b) If a and b are both real numbers, how can there be a imagionary number in between? where does imagionary numbers fit on the scale anywayz…. since it would be sqrt of -1, would it be inbetween 0 and -1?

    still, logically imagionary numbers seem to be a seperate catogory, so it wouldn’t even be on the number scale
    logically I agree with the problem, i dont see why you need to “prove it” cuz it’s already logical, (a) make sense, (b)… doesn’t

    RootBeer—–sqrt(BEER) = E sqrt(BR)
    lol, i wanna tell my joke in math club!

  4. paladin8 Says:
    November 23rd, 2006 at 4:20 pm

    b) It says irrational, not imaginary…

    And yes, continuity of f(x) = x^2 is a stronger result that the existence of square roots.

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