Quod Erat Demonstrandum Trivia Quiz

Answer: proof by induction

For instance, you want to verify that the sum of all numbers from 1 to n is (1/2)(n)(n+1). To do this you would need to verify that (1/2)(n)(n+1) + (n+1) = (1/2)(n+1)((n+1)+1). Distributing yields n²/2+n/2+n+1 = n²/2+3n/2+1. Multiplying both sides by two and combining like terms yields n²+3n+1=n²+3n+1, so the equation is true for all n.

For the incorrect choices:

Proof by contradiction is a proof where you start out by assuming the opposite of what you want to prove and then showing that that assumption leads to a ridiculous conclusion, and thus must be false.

Proof by contrapositive involves proving a different, yet equivalent statement. If you have a logical statement "If P, then Q", then its contrapositive is "If not Q, then not P". If one is true, then they both are. So if you are trying to prove a statement, you can opt instead to prove its contrapositive.

Proof by cases involves breaking up the problem into a finite number of cases which you can tackle individually. This is how the famous Four-Color Theorem was finally solved. Each map was put into one of several categories and then a computer was used to check that each category worked. Not all mathematicians were happy that it was proved this way, but almost all if not every mathematician trusts the result.

Answer: Derivating the function for a sphere of the same radius and evaluating it at the point r=0.

The integration method is best performed in Polar coordinates. ∫∫dA is the integral, where dA = r*drdθ. The bounds on r are [0,R] and the bounds on θ are [0,2π]. Performing the first step of the integral leaves you (1/2)∫R²*dθ. Evaluating this leaves A=πR².

The second method involves breaking up the circle into tiny sectors, with each sector's area approximately equal to r²sin(θ/2)cos(θ/2)=r²sin(θ)/2 (through the double angle identity) where r is the radius of the circle and θ is the central angle of the sector. If each sector has an angle θ, then the whole circle will contain a total of 2π/θ sectors. So the area will equal lim(2π/θ*r²sin(θ)/2) as θ→∞. Canceling the 2's and applying the limit product rule leaves you lim(πr²)*lim(sin(θ)/θ) as θ→∞. Since lim(sin(θ)/θ)=1, the limit evaluates to πr².

The third statement is just math gibberish.

Answer: Completing the Square

Completing the square:

1. ax²+bx+c = 0
2. x²+(b/a)x+(c/a) = 0
3. x²+(b/a)x = -c/a
4. x²+(b/a)x+b²/(4a²) = -c/a+b²/(4a²)
5. [x+b/(2a)]² = b²/(4a²)-4ac/(4a²)
6. [x+b/(2a)]² = (b²-4ac)/(4a²)
7. x+b/(2a) = ±√(b²-4ac)/(2a)
8. x = -b/(2a)±√(b²-4ac)/(2a)
9. x = [-b±√(b²-4ac)]/(2a)

"Negative b plus or minus the square root of b squared minus four a c all over two a."

The other choices:

Integration by parts is used to simplify a complicated integral into a simpler one.

Gaussian Elimination is used to simplify a matrix into reduced row echelon form.

Newton's method uses derivatives to find approximate answers to complicated equations.

Answer: If you assume there are finitely many primes, you can use that finite list of primes to find another prime.

If you have a set of all the primes (2,3,5,...,p) you can always find another prime by looking at the number you get when you multiply your finite list of primes and add one. Since dividing this number by any of the primes in your supposed complete list will always leave a remainder of 1, than this new number is either prime or prime factorable into two or more primes not on your list. Thus, give a finite set of primes, you can always find another one, so there are infinitely many primes.

A Mersenne prime is a prime number of the form 2ⁿ-1, but a number of that form is not necessarily prime (e.g. n=4). Prize money is awarded to people who can find primes over 10 million digits long. The largest known prime (when this quiz was made) is 2⁴³¹¹²⁶⁰⁹ - 1 and contains 12,978,189 digits. (It's a Mersenne prime.) Think you can find the next one?* Get to it!










* You won't.

Answer: the parallel postulate

Euclid's five postulate's were (generally speaking):

1. A line can be created by connecting two points.
2. Lines can be extended indefinitely.
3. A circle can be formed from a point [i.e. center] and a radius.
4. All right angles are congruent.
5. If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles [i.e. less than 180°], then the two lines intersect at some point.

Euclid could not prove this statement, but reasoned that it had to be true, so he left it as a postulate. Many mathematicians have attempted to prove this over the years, mainly because it is complex compared to the other four, which are very basic. It has not been proved, but a proof has been written proving that it cannot be proved one way or another. Two branches of geometry (hyperbolic and elliptical), both under the shell of non-Euclidean geometry were created, and Einstein actually discovered that when talking about very large spaces (like the universe), Euclidean geometry does not hold and in fact it is hyperbolic geometry that is more applicable. Cool, huh?

Answer: Step 5

The error in step 5 is that when dividing by a-b, we are really dividing by 0, because we set a = b = 1 at the beginning of the proof. Other than that one step though, all the other reasoning is sound.

It's a good thing that 0 ≠ 1 though, or we could prove anything we wanted to.

Try showing this "proof" to people and see if they can catch the error.

Answer: Step 6

The problem occurs in Step 6 because by subtracting S+1, I am subtracting infinity from infinity, which is a mathematical no-no. Up until that point, though, the proof is okay. 2S+1 = S is a perfectly valid equation. If you treat infinity as a number (it's okay here), then 2S+1 becomes 2*infinity+1 = infinity, so 2S+1 does in fact equal S. If this proof were correct, we would have such a ridiculous result. Imagine adding up an infinite string of positive integers and getting something negative!

Answer: If p is prime then, for any integer a, a^p = a (mod p).

This theorem is the motor behind the Fermat Primality Test. The Fermat Primality Test can say for certain that a number is not prime, but can only give a probability that something is prime. Say you want to see whether 319 is prime. Pick an a less than 319, say 101. So a=101 and p=319. 101^319 ≡ 182 (mod 319), so we can conclude that 319 is not prime (because our result was 182 and it should have been 101). It turns out that 11*29 = 319. Now let's test to see whether 317 is prime. We'll use 101 again. 101^317 ≡ 101 (mod 317). So far so good. Let's try another value for a, such as a = 251. 251^317 ≡ 251 (mod 317). Let's check one more, say a = 299. 299^317 ≡ 299 (mod 317). So we can conclude that 317 is probably prime, because Fermat's Test worked for three random, unrelated integers. It turns out that 317 is prime. However, it could have been the case that I was just lucky with my selections. There are times when a^p ≡ a (mod p) holds even when p is not prime. An example of this (from Wikipedia) is p=221 and a=38. The above equation is true even though 221 is composite. (It's 13*17.)

Note that if a is larger than p, you will have to take into account the fact that there exists b such that a ≡ b (mod p) and 0 ≤ b ≤ p. For instance, take p=2 and a=3. 3^2 ≡ 1 (mod 2). This may raise a red flag, but remember that 3 ≡ 1 (mod 2), so by the transitive property, 3^2 ≡ 1 ≡ 3 (mod 2).

For the other choices:

"If x, y, z, and n are natural numbers, there are no x, y, z such that x^n+y^n=z^n for n≥3." - This is Fermat's Last Theorem.

"If n≥2 and (n-1)! ≡ -1 (mod n), then n is prime." - This is Wilson's Theorem.

"If a and p are relatively prime, then a mod p ≠ 0." - This is a true statement though it's nothing earth-shattering.

Answer: Axiom of Choice

The Axiom of Choice (AoC) states that if you have a collection of sets (finite or infinite), it is possible to choose exactly one element from each set. Sounds reasonable, doesn't it? Well, if you assume it to be true, you can prove with absolute mathematical certainty the paradox stated in the question. Banach and Tarksi actually proved this theorem to try to contradict the AoC, but it turns out that by using the AoC to create the 5 or mores pieces, you are creating nonmeasurable sets, and thus the idea of volume doesn't really make sense, thereby eliminating any sort of logical contradiction. Most mathematicians accept the AoC because not only is it crucial in a number of important theories, but it also makes life easier in general for mathematicians trying to prove new theorems.

The other choices:

The Axiom of Infinity states that there exists at least one infinite set - the natural numbers.

The Axiom of Separation essentially states that subsets exists and are themselves sets.

The Axiom of Empty Set states that there exists a set with no elements - the empty set.

Answer: The hypothesis can neither be proved or disproved.

This was an incredibly startling conclusion. In 1931, Kurt Gödel proved the Incompleteness Theorem that implies (in a nutshell) that given any set of axioms, there are statements that can be neither proved nor disproved; however, until 1963, there had been no instances of this theorem appearing anywhere in math. The actual proofs of these theorems are difficult and therefore I won't attempt to go into them on this quiz.

I hope you enjoyed playing this quiz as much as I enjoyed writing it.