# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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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 , prove that there is a real number such that ). [Usual properties of the integers are assumed.]

Solution: Consider the set . We can show that is non-empty by selecting an integer large enough such that . Since is a real and the integers are unbounded, there exists a positive integer such that , thus so is bounded from above.

Now there must exist such that . We claim that . Suppose the contrary; then either or .

CASE 1: If , then consider . Choose large enough such that

which is definitely true for any . But then and , contradicting the fact that is the supremum.

CASE 2: If , then consider . Again, choose large enough such that

which we know is true for (furthermore, we impose the restriction so our resulting real is positive). Then for all and , contradicting the fact that is the supremum.

Hence we know that , 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 with , prove:

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

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

1. t0rajir0u Says:

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:

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:

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!