Is it True? Trivia Quiz

Answer: True

Assume there are a finite amount of primes. Thus, the primes can be listed as p1, p2, p3,...,pn. Consider the integer x = (p1*p2*p3*...*pn + 1). When x is divided by any prime, the remainder is one. Therefore, x must be prime. Thus, we prove by contradiction that there must be an infinite amount of primes.

Answer: False

This is a paradox made famous by Bertrand Russell. Consider S to be the set of all sets which do not contain themselves. If S is contained within S, then S contains itself, which implies that S does not contain itself. Similarly, if S is not contained within S, then S does not contain itself, which implies that S contains itself. Therefore, it is impossible for S to exist.

Answer: False

The general definition of conditional probability is P(x|y) = P(x&y)/P(y). The general rule for independence is P(x&y) = P(x)*P(y), which, when substituted into the conditional probability formula, yields P(x|y) = P(x) as another rule for independence. P(x|y) = P(x)*P(y) has no meaning in terms of the independence of x and y.

Answer: False

Let X be represented by some density function f(x). Also, let F(x) equal the cummulative distribution function of X. So, P(X=a) equals P(X is between a and a) which is equal to the integral of f(x) from a to a, which equals F(a) - F(a), which equals 0. Because the total number of real numbers is uncountable, it is impossible to have X equal an exact value, whether it be 3, 5/19, or 4.163454441009.

Answer: False

Let D and E represent the lengths of the legs of triangle ABC. Assume that both D and E are odd. If so, the let D = 2k + 1 for some integer k, and let E = 2j + 1 for some integer j. Let F equal the length of the hypotenuse. Using Pythagorean's Theorem, F^2 = (2k+1)^2 + (2j+1)^2 = (4k^2+4k+4j^2+4j+2) = 2*(2k^2+2k+2j^2+2j+1). Because k and j are integers, (2k^2+2k+2j^2+2j+1) must be odd because it is the sum of 4 even numbers and 1 odd number. So, F^2 = 2*n for some odd number n. So, F^2 is even, meaning it must also be the square of an even number.

However, all even perfect squares take the form (2m)^2, or 4m^2, where m is some integer, meaning that all even perfect squares must be multiples of 4. F^2 can only be a multiple 2, because q cannot have 2 as a factor since it is odd.

Therefore, F cannot be an integer if D and E are odd, and triangle ABC cannot be a right triangle if D and E are odd.

Answer: True

If Q(n) implies Q(n+2), then Q(0) implies that Q(2) is true, which implies that Q(4) is true, which would eventually prove that Q(n) is true for all even numbers. Also, if we look at the contrapositive of Q(n) implies Q(n+2), then we know that if Q(n+2) is false, then Q(n) is false. So, since Q(1001) is false, Q(999) is false, Q(997) is false, and so Q(n) is false for all odd integers less than or equal to 1001. So, since Q(n) is true for all even integers between 1 and 1000 and all odd integers between 1 and 1000, we know whether or not Q(n) is true for all integers between 1 and 1000.

Answer: False

Consider the probability that nobody in a group of people have the same birthday. If there is one person in the group, then the probaility that he doesn't share a birthday with anyone is 365/365. If you add another person to the group, the probability that he doesn't share a birthday with anyone is 364/365. Add another person, and the probability is 363/365. So, for a group of people, the probability that no one shares the same birthday is 365/365 * 364/365 * 363/365 * ... * (365-n+1)/365, or (365 P n)/365^n for a group of n people. So, the probability that at least two people have the same birthday is 1 - (365 P n)/365^n.

In a group of 50 people, the probability that at least two people share a birthday is about 0.97. For 23 people, the probability is about 0.5.

Answer: True

Let the length of one leg be represented by x, and the length of the other be represented by k*x, where k is some integer greater than 0. Also, let the length of the hypotenuse be represented by y. Using Pythagorean's Theorem, we know that:

x^2 + (k*x)^2 = y^2
x^2 + k^2 * x^2 = y^2
x^2*(k^2 + 1) = y^2

Since the triangle has all integer side lengths, y^2 must be a perfect square, which means x^2*(k^2 + 1) must be a perfect square, which means k^2 + 1 must be a perfect square. However, since k is an integer greater than 0, and the difference between any two consecutive perfect squares greater that 0 is at least 3, k^2 + 1 cannot be a perfect square. Thus, it is impossible for the lengths of the legs of a right triangle to be multiples of one another if they are integers.

Answer: True

If a number in base 10 ends in 111, then it must take the form (1000k + 111) for some non-negative integer k. If a number in base 2 ends in 111, then it must take the form (8j + 7) in base 10 for some non-negative integer j. In other words, if a number in base 10 has a remainder of 7 when divided by 8, it will end in 111 in base 2.

When we divide (1000k + 111) by 8, we get (125k + 111/8). 111/8 = 13 + 7/8, so when (1000k + 111) is divided by 8, it has a remainder of 7, meaning that the binary representation of any base 10 number ending in 111 must also end in 111.

Answer: False

When infiinity is substituted for n, the limit becomes infinity/infinity, so we must use L'Hopital's rule (derivative of the numerator and denominator) to find the limit. However, since k can be any integer greater than 1, we must use induction to show that the limit will always be infinite. First, consider k = 2. Using L'Hopital's rule, we show that the limit of (e^n/n^2) as n tends to infinity is equals to the limit of (e^n/2n). Using L'Hopital's rule again, we get the limit of (e^n/2), which is infinite. Now, consider some integer q. Assume that (e^n/n^q) tends to infinity as n tends to infinity.

Then, the limit of (e^n/n^(q+1)) as n tends to infinity is equals to the limit of (e^n/[(q+1)*n^q]), based on L'Hopital's rule. That limit is equal to 1/(q+1) * lim as n tends to infinity of (e^n/n^q), which, as previously stated, tends to infinity. So, since (e^n)/(n^k) tends to infinity when k = 2, and if (e^n)/(n^q) tending to infinity implies that (e^n)/(n^(q+1)) tends to infinity, then we can say that for all integers k > 2, the limit as n approaches infinity of (e^n)/(n^k) is infinite, and never a constant.