That Can't Be Right! Trivia Quiz

Answer: False

Take, for example, the set E ⊂ Q (Q denotes the rational numbers) of all x such that 2 ≺ x² ≺ 3. E is open because every point is an interior point; there is no rational number whose square is 2 or 3, so for any x ∈ E you choose, there is guaranteed to be at least one point on each side of it that is also in E. E is closed because it contains all of its limit points; √2 and √3 are not rational numbers and thus are not limit points. If E were a subset of the real numbers, though, it would not be closed because √2 and √3 would be limit points not contained in the set.

A set that is both open and closed is sometimes referred to as a "clopen" set.

Answer: False

It's actually just less than 8%. While it is the most probable outcome, it is still not very likely. This is a binomial probability problem. The answer is given by nCr(100,50)*0.5^50*0.5^50 = 7.95%.

You may be wondering why I said you wouldn't need a calculator for this quiz. I stand by my statement, though. While it would be difficult without a calculator to find that the probability was 7.95% or even that it was less than 10%, your intuition should have told you that 25% was way too high.

Answer: False

The best you can do is a circle.

Note that while you need calculus to prove this, you do not need calculus to show that there exists a shape that will give you more area than a square.

If you are given a f feet of fencing, and you want to construct from that a regular polygon with n sides, each side thus has length f/n. The area of a regular polygon is given by (1/2)ap, where a is the apothem (the perpendicular distance from the center to one of the sides) and p is the perimeter.

With some simple trigonometry, you can find that the apothem is equal to f/(2n*tan(π/n)). Thus the area of a regular n-gon with perimeter f is equal to A = f²/(4n*tan(π/n)). Since f is fixed, it is enough to find the minimum value of 4n*tan(π/n) (thus maximizing A). Looking at the derivative of 4n*tan(π/n), you will find that it approaches 0 as n approaches infinity. Thus n=∞ will give you the minimum value of 4n*tan(π/n).

Thus, given a fixed amount of fencing, the shape containing the most area is an infinite-sided polygon, or in other, more familiar terms, a circle.

Note that if you take the limit of f²/(4n*tan(π/n)) as n approaches infinity, you will get f²/(4π), which is exactly the area of a circle with circumference f. Why? Since the circumference of the circle equals 2πr = f, r = f/(2π). Thus the area, πr², equals π[f/(2π)]² = f²/(4π).

Answer: False

f is a constant function: f(x) = c.

Proof:

Since (x-y)² is always positive, this is the same as |x-y|², which is the same as |x-y|*|x-y|.

Dividing both sides by |x-y| yields: |f(x)-f(y)|/|x-y| ≤ |x-y|.

Taking the limit of both sides as x approaches y and recognizing that the right side of the equation is now f '(x), we have |f '(x)| ≤ 0. Thus, f '(x) = 0, implying that f(x)=c, for some c.

Answer: False

Let at a = b = √2. (√2)^(√2) is either rational or irrational. If it is rational, then we have just found a and b that contradict the statement. If it is irrational, let a = (√2)^(√2) and b = √2. Thus, a^b = [(√2)^(√2)]^(√2) = (√2)² = 2.

Thus, there exist irrational a and b such that a^b is not irrational.

Answer: False

In math terms, this statement is saying that this series converges to a number strictly less than 1. (I will now show that this statement is false.)

Do not confuse this with the harmonic series, which we know diverges.

Let's look at the first few terms. The first time, you walk 1/2*1/1. The second time, you walk 1/3*1/2. The third time, you walk 1/4*(1-1/2-1/6) = 1/4*1/3. The fourth time, you walk 1/5*(1-1/2-1/6-1/12) = 1/5*1/4. Do you see the pattern developing?

This is equivalent to the sum 1/[n(n+1)], as n goes from 1 to infinity. However, if you rewrite the series as the sum of 1/n - 1/(n+1) as n goes from 1 to infinity, it will be easier to see why this converges. Write out the first few terms:

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - ... - 1/n + 1/n - 1/(n+1).

You can see that except for the first and last terms, every term cancels out with the previous term. A series that does this is called a telescoping series.

Now, your sum is just a simple limit. Namely, the limit of 1-1/(n+1) as n goes to infinity, which is clearly 1.

Thus, after repeating the process infinitely many times, you will reach you destination. (The time it takes to complete each step approaches 0 though, so you will reach your destination in finite time. Otherwise we'd never get anywhere!)

Answer: False

It misses the point 0, so its range is actually (-∞, 0) ⋃ (0, ∞). e^z = a+bi always has a solution, unless a=b=0. In fact, z = ln(a+bi) = ln|a+bi| + arg(a+bi), where |a+bi| is the norm and arg(a+bi) is the angle made with the positive real axis.

Answer: False

It actually equals 1. (How cool is that!?) The reason is basically because for 0 ≺ x ≺ 1, -x^n approaches 0 so e^(-x^n) approaches 1. For 1 ≺ x ≺ ∞, -x^n approaches -∞, so e^(-x^n) approaches 0. Thus, your function can be approximated by the very simple piecewise function: f(x)=1 for 0 ≺ x ≺ 1 and f(x)=0 for 1 ≺ x ≺ ∞. Calculating the integral of this is very easy and boils down to finding the area of a one-by-one square, which is, of course, 1.

Another interesting tidbit is that for all finite n, e^(-x^n) is a continuous function on [0, ∞). However, taking the limit as n goes to infinity gives this function a simple discontinuity at x=1, where f(x)=1/e.

And while I'm thinking of the value 1/e, let me just mention that it happens to be the value of x that produces the minimum value of the function f(x) = x^x.

Answer: False

The equation you need to solve is actually x² - tr(A)x + det(A) = 0. Let the matrix A equal:
[a b]
[c d]

Solving the equation used to find the eigenvalues, det(xI-A) = 0 (where I is the identity matrix), you get:
|x-a -b|
|-c x-d| = 0

Using the formula for det(A), ad-bc, we have (x-a)(x-d) - (-b)(-c) = x² - (a+d)x + (ad-bc) = 0.

Notice the coefficient on x and the constant term are just tr(A) and det(A), respectively.

Thus, the eigenvalues of A are the solutions to the equation x² - tr(A)x + det(A) = 0.

Answer: False

Just like in Question 4, this function is also a constant (though for a very different reason, one that involves Cauchy's integral formula and the Taylor series of a complex function). This is known as Liouville's Theorem, named after the French mathematician Joseph Liouville who proved this.

You may be saying, "Wait! |sin(z)| ≤ 1 and |cos(z)| ≤ 1, so why isn't Asin(z)+Bcos(z) bounded?" The reason is that when the domain is extended to the complex numbers, sin(z) and cos(z) lose their property of being bounded functions, and both approach infinity as z approaches infinity.

----------------------------------------------

You may have noticed that all of the answers were False, but hey, the name of the quiz was "That Can't Be Right!".

Thanks for playing.