Jeffrey Shallit posted the questions for the 2010 Bernoulli Trials at the University of Waterloo (Waterloo, Ontario, Canada). This is a math contest for undergraduates. You have to decide whether each of following questions are true or false.

Professor Shallit liked the questions because one or two made him think for a bit. Bully for him! When I looked at the list I can honestly say that none of them (not a single one) made me think of anything except what is a Bernoulli and why is he doing this to undergraduates?

I wonder if I could made up questions like this for biology students? I don't think so—the choices would have to be "mostly true" and "mostly false."

I think I got #1 and #3. Other than that....

ReplyDeleteI think 4 is false (1234567879 contain 9 numbers, then the numbers 10-19 contain 10 numbers =/= 1/9), 10 is false (it's too specific), and 11...I'm not sure about but intuitively I would say its false.

ReplyDeleteI'm sure that Prof. Moran is well aware that the Bernoulli brothers were naturalists who were contemporary with the most important scientist who ever lived, Issac Newton. As he was wont to do, Newton carried on a feud with them, as he did with virtually every other contemporary naturalist.

ReplyDeleteDifficult. Trying #1.

ReplyDelete(x + 1/x) = n, x not equal +/- 1.

square both sides

(x + 1/x)^2 = n^2

L.H.S. = (x + 1/x)(x + 1/x)

= x^2 + 1 + 1 + 1/x^2

= x^2 + 1/x^2 + 2

= R.H.S

= n^2

Take square root of both sides

n = +/-(x + 1/x + sqrt(2))

which is irrational. So, nope, no integer that (x + 1/x) = n.

That work?

n = 1: z^1 = x^2 + y^2 is true.

ReplyDeletelet z = 2, x = 1, y = 1. (2 = 1 + 1)

n = 2: z^2 = x^2 + y^2 is true.

let z = 10, x = 6, y = 8 (100 = 36 + 64)

n = 3: z^3 = x^2 + y^2 is true.

let z = 2, x = 2, y = 2 (8 = 4 + 4)

Perhaps induction can be used?

P(k): z^n = x^2 + y^2 where n,x,y,z are positive integers.

P(1) is true.

P(k) = z^k = x^2 + y^2

P(k+1) = z^(k + 1) = x^2 + y ^ 2.

= z(z^k) = x^2 + y^2.

As z is an integer and z^k is an integer (integer to power of some integer) then P(k+1) is true. There exist positive integers x,y,z such that z^n = x^2 + y^2 is true for each positive integer n.

Does that work?

Regarding 15, how can a rectangle have 3 distinct side lengths? Seems illogical.

ReplyDeleteA short guide to the Bernoullis by your truly ;)

ReplyDeleteBrian:

ReplyDelete+/-(x + 1/x + sqrt(2))

is not the sqrt of

x^2 + 1/x^2 + 2

(I think #1 is false, but mostly by intuition)

I don't think this works:

ReplyDeleteP(k) = z^k = x^2 + y^2

P(k+1) = z^(k + 1) = x^2 + y ^ 2.

= z(z^k) = x^2 + y^2.

In induction, you have to prove that IF the proposition is true for P(k) THEN it is also true for P(k+1). You have simply assumed that there exist x and y such that P(k=1) is true -- but they are presumably not the same as the x and y for P(k). Something has to happen to the RHS to demonstrate that the identity still works.

1) is false.

ReplyDelete1) x + 1/x = N ( N is an iNteger)

multiplying both sides by x gives:

2) x^2 +1 = Nx

rearranging:

3) x^2 -Nx +1 = 0

solving by quadratic formula:

4) x = ( N +/- root(N^2 -4) ) / 2

since x, N, 2 are all rational, root(N^2 -4) must also be rational

this is trivially true when N=2 or N=-2 and correspondingly x=1 or x=-1.

N is an integer, and root(N^2-4) is imaginary for values of N = -1, 0 or +1.

all you have to do then is prove that root(N^2 -4) is irrational for all whole number values of N =3 or larger. :)

Thanks Eamon. Guess I need to study more maths. :)

ReplyDelete

ReplyDeleteall you have to do then is prove that root(N^2 -4) is irrational for all whole number values of N =3 or larger.Apparently, Dedekind and others have well-established that the square roots of non-square natural numbers are irrational. See: http://en.wikipedia.org/wiki/Quadratic_irrational for one such proof.

Since the distance between perfect squares of size 3^2 or larger is >4, N^2-4 is not a perfect square for all values of N >=3, therefore 1 + 1/x has no integer solutions for rational values of x other than 1, -2. QED :)