Math Trivia 3 Trivia Quiz

Answer: 1

25 = 3*8 + 1, so the quotient is 3 and the remainder is 1.

Answer: (x - 10)(x + 10)

x^2 - 100 is a difference of squares since 100 = 10^2. Recall that a difference of squares factors as follows:

A^2 - B^2 = (A - B)(A + B)

So x^2 - 100 = (x - 10)(x + 10).

Answer: 20

The third side of a triangle is greater in length than the difference of the other two sides and smaller in length than the sum of the other two sides. So for this triangle, let x denote the length of the third side. It satisfies:

10 - 8 is less than x is less than 10 + 8

or 2 is less than x is less than 18.

The only value of x I gave that doesn't work is 20.

Answer: The hyperbola

An asymptote can be thought of as a line the curve approaches. As an easy example, consider the hyperbola y = 1/x. As x goes to infinity (or -infinity), y goes to 0. So the line y = 0 is a horizontal asymptote. As x approaches 0 from the right, y goes to infinity. So the line x = 0 is a vertical asymptote. None of the other three conic sections listed have asymptotes.

Answer: It's a constant.

Any two antiderivatives of a continuous function differ by a constant. Hence the need for the "+ C" in indefinite integrals.

Answer: 0

The cross product of two non-zero vectors a and b in R^3 is orthogonal (perpendicular) to both a and b. So a and c are orthogonal. The dot product of two orthogonal vectors is 0.

Answer: All of these

All of these are true:

Let x be an element of A - B. Then x is an element of A that's not in B, but in any case x is an element of A. So (A - B) is a subset of A.
To see that (A - B) and B are disjoint, let x be an element in their intersection. Then x is in (A - B), so x is in A but not in B. x is also in B. So x is in B and x is not in B. Contradiction, so the intersection is empty.
Now let y be an element of A that is not in the intersection of A and B. Then y is in A but not in B, so y is in A - B, therefore the union of (A - B) and (A intersect B) is A. You can also see that all these are true by drawing a Venn diagram.

Answer: NOT( (NOT p) OR (NOT q) )

By one of DeMorgan's laws, NOT(p OR q) is logically equivalent to (NOT p) AND (NOT q). Therefore,

NOT( (NOT p) OR (NOT q) )

is logically equivalent to (NOT( NOT p )) AND (NOT( NOT q ))

which is logically equivalent to p AND q.

What this means is we can get by with just OR and NOT, we technically don't need AND, but it is nice of course to have both AND and OR. By the way, DeMorgan's other law states that NOT(p OR q) is logically equivalent to (NOT p) AND (NOT q), which means we could also get by with just AND and NOT.

Answer: 3v

If Ax = (lambda)*x, we say that lambda is an eigenvalue of A and x is an eigenvector of A belonging to lambda. So this question is about unraveling the definition of eigenvalue and eigenvector. These notions are extremely important in linear algebra.

Answer: (10 choose 6)

This is from Pascal's identity, which shows how to get successive rows from the previous rows in Pascal's triangle:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
.
.
.

Let me show you directly why the following sum is correct. Consider a ten element set. There is more than one way to compute the number of 6 element subsets of it. The number of 6 element subsets of this set is just (10 choose 6). But here's another way to count: Let a denote an element of this set. Then either a 6 element subset contains a or it doesn't. The number of 6 element subsets that contain a is (9 choose 5) - we need to select 5 elements that are not equal to a from the remaining 9 elements. The number of 6 element subsets that don't contain a is (9 choose 6) - just select 6 elements that aren't a from the 9 that aren't a. So adding these two numbers gives the number of 6 element subsets of a ten element set, which is just (10 choose 6).

Answer: 3

Any integer is of the form 2n or 2n + 1. Now look at their squares:

(2n)^2 = 4n^2, (2n + 1)^2 = 4n^2 + 4n + 1 = 4(n^2 + n) + 1.

So if x is an integer, x^2 is of the form 4k or 4k + 1 for some integer k. This means that x^2 is either congruent to 0 or 1 (mod 4). So if x and y are integers, x^2 + y^2 is congruent to 0, 1, or 2 (mod 4). Thus x^2 + y^2 cannot be congruent to 3 (mod 4).

Answer: 1

Let S denote the set of all possible limits of subsequences of a_n. The infimum (or greatest lower bound) of S is 2 - 1 = 1, which represents the limit of the subsequence 2 + (-1)^(2n + 1). Thus 1 is the lim inf of the sequence. Note that the limit of this sequence doesn't exist, but the lim inf and lim sup both exist (the lim sup is 3).

Answer: The cyclic group of order 6, Z6

Abelian means the multiplication is commutative. All cyclic groups are abelian groups. Cyclic means the group has one generator. So let G be a cyclic group generated by the element a. If x,y are elements of G, then x = a^m, y = a^n for some integers m and n. (If G were finite, m and n would necessarily be non-negative). Then

xy = (a^m)(a^n) = a^(m + n) = a^(n + m) = (a^n)(a^m) = yx.

The cyclic group of order 6 is isomorphic to Z/6Z (The integers mod 6 under addition). It is also isomorphic to Z/2Z x Z/3Z.

The other answers are classic examples of groups, and I encourage you to find out more. The symmetric group S4 has 24 elements, each of which is a permutation on 4 letters. The dihedral group D8 consists of rigid symmetries of the square. Finally, the quaternion group Q consists of eight elements:

1, -1, i, -i, j, -j, k, -k

where i^2 = j^2 = k^2 = -1

and ij = k, jk = i, ki = j, ji = -k, kj = -i, ik = -j.

Answer: x^2 + 4xy + y^2

(x + y)^2 = (x + y)(x + y)

= x^2 + xy + yx + y^2

= x^2 + xy + 3xy + y^2 (since yx = 3xy)

= x^2 + 4xy + y^2.

Let me now show you that this can actually happen. Let R be the ring of 2 by 2 matrices with entries in the set of real numbers. Let x be the 2 by 2 matrix with a 1 in row 1, col 2 and zeros elsewhere. Let y be the 2 by 2 matrix with 3 and 1 on the diagonal, zeros elsewhere. Then xy = 3yx.

Answer: The inverse image of V under f is an open set of X

This is precisely the definition of a continuous function from a topological space X to a topological space Y. When X is the set of real numbers, this is much neater than the usual "epsilon-delta" definition. Note that in general the continuous image of an open set need not be open. As an easy example, let X and Y both be the set of real numbers and let f be the continuous function that takes any real number and maps it to 0. Then f maps each open set in X to the closed set {0}.

I hope you enjoyed this quiz! Thanks for playing!