Math Trivia 1 Trivia Quiz

Answer: 17

By order of operations, you need to do the multiplication first, then the addition:

3 + 2*7 = 3 + 14 = 17

Answer: 4x^2 - 28x + 49

Recall that (A - B)^2 = A^2 - 2AB + B^2. Put A = 2x, B = 7:

(2x - 7)^2 = (2x)^2 - 2*(2x)*7 + 7^2 = 4x^2 - 28x + 49.

Answer: The circumcenter

The circumcenter is the point of the triangle that is equidistant from each of the three vertices. The perpendicular bisector of a side represents the set of points that are equidistant from two of the vertices, so where two perpendicular bisectors intersect, we find a point that is equidistant from all three vertices. It's interesting that the three lines are concurrent, meaning that they all pass through the same point. The circumcenter is the center of a circle that passes through all three vertices of the triangle (also known as the "circumcircle").

The angle bisectors are concurrent in the incenter, the medians are concurrent in the centroid, and the altitudes are concurrent in the orthocenter.

Answer: e is greater than 1

All conic sections can be described in terms of a fixed point, called a focus, and a fixed line, called a directrix. Let P be a point on the conic, let F be the focus, and let l denote the directrix. Then the eccentricity e can be defined as the following ratio:

e = (the distance from P to F)/(the distance from P to l)

This ratio in fact is always a constant for a conic section. We can use the concept of eccentricity to distinguish all the conic sections. If e is greater than 1, the conic is a hyperbola. If e = 1, the conic is a parabola. If e is between 0 and 1, the conic is an ellipse. Finally, we will define the eccentricity of a circle to be 0. The eccentricity can also be computed from the standard form of the equations of an ellipse or hyperbola.

Answer: f(x) has a local minimum at x = 4

Since f '(x) is negative for all x less than 4, f(x) is decreasing for all x less than 4. Since f '(x) is positive for all x greater than 4, f(x) is increasing for all x greater than 4. Therefore, f(x) has a local minimum at x = 4 by the first derivative test.

The first derivative test will determine the nature of a local extrema even when the second derivative test fails. So in that respect, it is a better test. However, the second derivative test is faster when the second derivative is easier to compute. Note that we couldn't use the second derivative test here since I gave no information about it (in fact, it didn't have to exist anywhere!)

Answer: The vector (yz, xz, xy)

The components of the gradient are the partial of f with respect to x, the partial of f with respect to y, and the partial of f with respect to z. Partial derivatives are found by differentiating with respect to the variable, treating the other variables as constants.

Answer: 6

The number of 2 element subsets of a 4 element set is just C(4,2) = 6. You can also list them: Suppose S = {a, b, c, d}. These are the two element subsets of S:

{a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}

Answer: nilpotent

Nilpotent matrices are important in the study of canonical forms. An orthogonal matrix is one whose transpose is equal to its inverse, a symmetric matrix is one that equals its transpose, and an idempotent matrix is one that equals its square.

Answer: 17

Let a and b be integers. We say that a is congruent to b mod n if n divides a - b, in other words, if there exists an integer k such that

a - b = kn.

17 is the answer because when we divide 17 by 7, the remainder is 3:

17 = 2*7 + 3 so 17 is congruent to 3 mod 7.

You can check that mod 7, 27 is congruent to 6, 37 is congruent to 2, and 47 is congruent to 5.

Answer: 1/6

I'll give two different solutions.
Solution 1: The probability the first bulb selected is good is 2/4 = 1/2. Now out of the three bulbs remaining, only 1 is good. So the probability that the second is good given that the first is good is 1/3 (note this is a conditional probablility). Therefore, the probability that the first bulb is good and the second bulb is good is found by:

(1/2) * (1/3) = 1/6.

Solution 2: The sample space has C(4,2) elements (it's the number of ways we can select 2 light bulbs from the 4 in the box). We need the number of ways we can select exactly 2 good bulbs: We can select 2 good bulbs from the 2 good bulbs C(2,2) ways, and we can select 0 bad bulbs from the 2 bad bulbs C(2,0) ways. Hence the probability is given by:

C(2,2) * C(2,0)/C(4,2) = 1*1/6 = 1/6.

Note that I include the extra term in the numerator to emphasize this is a hypergeometric probability (marbles in urns, committee problems, etc. have probabilities that take this form).

Answer: f is differentiable at some value c in the interval

Continuous functions do not have to be differentiable anywhere! For a proof of the existence of a nowhere differentiable function on the interval [0,1], see James R. Munkres, "Topology", second ed., section 49. In fact, he proves something better: Given any continuous function f on [0,1] and any epsilon greater than 0, there exists a continuous function g on [0,1] with

|f(x) - g(x)| less than epsilon for all x

and g is nowhere differentiable.

Other answers: The fact that f attains its maximum value at some c in [0,1] is the extreme value theorem. Continuous functions are Riemann integrable. Finally, [0,1] is a closed and bounded set of real numbers, hence is compact by the Heine-Borel Theorem. Therefore, the image of f is compact (the continuous image of a compact set is compact).

Answer: The center of G

The center of G is in fact a subgroup: Let x, y be elements of the center, and let g be an element of G. Then

(xy)g = x(yg) by associativity

= x(gy) since y is an element of the center

= (xg)y by associativity

= (gx)y since x is an element of the center

= g(xy) by associativity. Thus xy is in the center of G.

Also, x^(-1)g = (g^(-1)x)^(-1) by the formula for the inverse of a product

= (xg^(-1))^(-1) since x is in the center of G and g^(-1) is an element of G

= gx^(-1) hence x^(-1) is in the center. Therefore, the center of G is a subgroup. Note that G doesn't have to be finite.

Other answers: The commutator subgroup of G is the subgroup generated by elements of the form xyx^(-1)y^(-1) for x,y elements of G. The trivial subgroup is the subgroup consisting only of the identity element. Finally, if p is a prime such that p^k divides the order of G but p^(k+1) doesn't divide the order of G, then a subgroup of order p^k is called a Sylow p-subgroup. Sylow p-subgroups exist by the famous "Sylow Theorems." Moreover, any two Sylow p-subgroups are conjugate and the number of Sylow p-subgroups is congruent to 1 mod (order of G divided by p^k).

Answer: There exists an ideal J not equal to R or m with m a subset of J a subset of R

Since m is maximal, there does not exist an ideal between it and R (with respect to inclusion). Note that m maximal implies that R/m is a field implies that R/m is an integral domain (fields are integral domains) which implies that m is a prime ideal (p is a prime ideal of R iff R/p is an integral domain).

I realize there's a lot of terminology here, but I'll give you a little idea of what this is about:

A ring R is a set with two operations: addition and multiplication. The sum of two elements of R is another element of R, and the product of two elements of R is another element of R. This ring R is commutative, which means the multiplication satisfies xy = yx for all x,y in R. It has 1 means there exists an element 1 in R such that x*1 = x for all x in R. The elements of a ring also satisfy a number of axioms, but they are all "logical" and what you would want to be true.

An integral domain is a commutative ring with 1 with the additional property that whenever a,b are elements of R such that ab = 0, then a = 0 or b = 0. This should look very familiar, since the set of real numbers is an integral domain.

A field has the additional property that every nonzero element has a multiplicative inverse.

Answer: e^(i*pi/2)

Recall Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta).
The sixth roots of 1 are:

e^(2*pi*i/6) = cos(2*pi/6) + i*sin(2*pi/6) = (1/2) + i*(sqrt(3)/2)

e^(4*pi*i/6) = cos(2*pi/3) + i*sin(2*pi/3) = -(1/2) + i*(sqrt(3)/2)

e^(6*pi*i/6) = cos(pi) + i*sin(pi) = -1

e^(8*pi*i/6) = cos(4*pi/3) + i*sin(4*pi/3) = -(1/2) - i*(sqrt(3)/2)

e^(10*pi*i/6) = cos(5*pi/3) + i*sin(5*pi/3) = (1/2) - i*(sqrt(3)/2)

e^(12*pi*i/6) = cos(2*pi) + i*sin(2*pi) = 1

The incorrect answer e^(i*pi/2) = cos(pi/2) + i*sin(pi/2) = i. Note that i^6 = i^2 = -1, so i is not a sixth root of 1.

Answer: It must be open

A compact subset of a Hausdorff space is closed, hence its complement is open. There is a lot of terminology here to digest:

First of all, a topology on a set X is a collection of sets, called open sets with the property that any union of open sets is open, any finite intersection of open sets is open, and X together with the empty set is open. The set X together with a topology on X is called a topological space.

A subset C of X is compact if whenever C is contained in a union of open sets, a union of a finite subcollection of those open sets contains C (we say that every open cover of C has a finite subcover).

A topological space is Hausdorff if whenever x and y are two points in X, there exists disjoint open sets U and V such that x is in U and y is in V.

Finally, closed sets are just complements of open sets.

For a proof of the fact that a compact subset of a Hausdorff space is closed, see Theorem 26.3 on p. 165 of "Topology", second ed., by James R. Munkres.

I hope you enjoyed this quiz! I wanted to get across just how big the subject of mathematics actually is, but this quiz really only gives a small taste. It seems that many people believe there is nothing beyond calculus, when in fact the truth is math really begins with calculus! The other truth is that there is more to math than calculus - algebra, topology, logic, ... and that all of mathematics is not linear, it has many different branches. Thanks for playing!