Proofs

The proof that the square root of 5 is irrational is exactly the same as the well-known proof that the square root of 2 is irrational - except using 5 in place of 2. We can prove a more general result: the square root of any prime is irrational.

First of all, we require the lemma:

for any prime p, and integer x,

p|x2 ⇒ p|x

That is, if x2 is divisible by p, then so is x.

Proof:

The prime factorization of x2 necessarily contains p at least once, since it is divisible by p. But it also has to contain an even power of every prime, since it is the prime factorization of a square. Therefore, it contains p at least twice, and its square root, x, contains p at least once: that is, x is divisible by p.

Now, given a prime p, assume that its square root is rational. Then, it may be written in the form a/b, where a and b have no common factors (that is, the fraction a/b is in lowest terms). This is always possible for any nonzero rational number. Since this quantity is the square root of p, its square equals p, that is

(a/b)2 = p

a2/b2 = p

a2 = pb2

Now, pb2 is a multiple of p, so a2 must be too. And, using the result above, this means that a must be a multiple of p also. Thus, there exists an integer c such that

a = PC

Then,

(PC)2 = pb2

p2c2 = pb2.

Since p is not zero, we may divide both sides by p to obtain

PC2 = b2

That is, b2 is divisible by p also, and thus b is divisible by p.

Since a and b were both divisible by p, the fraction a/b could not have been in lowest terms, which contradicts our initial assumption. Therefore, the square root of p cannot possibly be a rational number. Since 5 is prime, the proof is complete.

Yes.

Although the definition of a parallelogram is "a quadrilateral with both pairs of opposite sides parallel", the only way for a quadrilateral to include opposite sides of equal length is if the included angles are the same, and hence the sides are parallel.

(Hint : draw a diagonal to a parallelogram. You can show that one of the two triangles formed is the mirror image of the other, which immmediately proves that each pair of opposite sides is equal.)

Given a three digit number n = "d1 d2 d3" (i.e. a number 0 <= n <= 999, where d1, d2 and d3 are it's digits), we can express n as 100d1 + 10d2 + 1d3. The number n' with the digits of nin reverse order would be: n' = "d3 d2 d1", which could be expressed as n' = 100d3 + 10d2 + 1d1. Subtracting n from n' (or vice versa) we get the following equation:

n - n' = (100d1 + 10d2 + 1d3) - (100d3 + 10d2 + 1d1)

resolving and rearranging we obtain:

n - n' = (100d1 - 1d1) + (10d2 - 10d2) + (100d3 - 1d3) = 99d1 - 99d3 = 99(d1 - d3)

Since d1 and d3 are integers (and d2 cancels out), we see that the result 99(d1 - d3) is always divisible by 99.

(The equation holds for all integers values for d1, d2 and d3, not just for the digits 0, 1, ... , 9, but we can no longer write it as a "three digit number". )

Example:

651 - 156 = (6*100 + 5*10 + 1*1) - (1*100 + 5*10 + 6*1) = 500 - 0 - 5 = 495 = 99*5

Given a matrix A=([a,b],[c,d]), the trace of A is a+d, and the det of A is ad-bc.

By using the characteristic equation, and representing the eigenvalues with x, we have the equation

x2-(a+d)x+(ad-bc)=0

Which, using the formula for quadratic equations, gives us the eigenvalues as,

x1=[(a+d)+√((a+d)2-4(ad-bc))]/2

x2=[(a+d)-√((a+d)2-4(ad-bc))]/2

now by adding the two eigenvalues together we get:

x1+x2=(a+d)/2+[√((a+d)2-4(ad-bc))]/2+ (a+d)/2-[√((a+d)2-4(ad-bc))]/2

The square roots cancel each other out being the same value with opposite signs, leaving us with:

x1+x2=(a+d)/2+(a+d)/2

x1+x2= 2(a+d)/2

x1+x2=(a+d)

x1+x2=trace(A)

Q.E.D.

The proof uses the following ingredients:

(1) Every nxn matrix is conjugate to an upper-triangular matrix

(2) If A is upper-triangular, then tr(A) is the sum of the eigenvalues of A

(3) If A and B are conjugate, then tr(A) = tr(B)

(4) If A and B are conjugate, then A and B have the same characteristic polynomial (and hence the same sum-of-eigenvalues)

If these are all true, then we can do the following: Given a matrix A, find an upper-triangular matrix U conjugate to A; then (letting s(A) denote the sum of the eigenvalues of A) s(A) = s(U) = tr(U) = tr(A).

Now to prove (1), (2), (3) and (4):

(1) This is an inductive process. First you prove that your matrix is conjugate to one with a 0 in the bottom-left corner. Then you prove that this, in turn, is conjugate to one with 0s at the bottom-left and the one above it. And so on. Eventually you get a matrix with no nonzero entries below the leading diagonal, i.e. an upper-triangular matrix.

(2) Suppose A is upper-triangular, with elements a1, a2, ... , an along the leading diagonal. Let f(t) be the characteristic polynomial of A. So f(t) = det(tI-A). Note that tI-A is also upper-triangular. Therefore its determinant is simply the product of the elements in its leading diagonal. So f(t) = det(tI-A) = (t-a1) * ... * (t-an). And its eigenvalues are a1, ... , an. So the sum of the eigenvalues is a1 + ... + an, which is the sum of the diagonal elements in A.

(3) This is best proved using Summation Convention. Summation convention is a strange but rather useful trick. Basically, the calculations I've written below aren't true as they're written. For each expression, you need to sum over all possible values of the subscripts. For example, where it says b_ii, it really means b11 + b22 + ... + bnn. Where it says bil deltali, it means (b11 delta11 + ... + b1n deltan1) + ... + (bn1 delta1n + ... + bnn deltann). Oh, and deltakj=1 if k=j, and 0 otherwise.

Suppose B = PAP-1. Let's say the element in row j and column k of A is ajk. Similarly, say the (i,j) element of P is pij, the (k,l) element of P-1 is p*kl, and the (i,l) element of B is bil. Then:

bil = pij ajk p*kl

And the trace of B is given by:

tr(B) = bii

= bil deltali

= p*kl deltali pij ajk

= p*kl plj ajk

= deltakj ajk (since p and p* are inverses)

= ajj

= tr(A)

(4) Again, suppose B = PAP-1. Then, for any scalar t, we have tI-B = P(tI-A)P-1. Hence det(tI-B) = det(P).det(tI-A).det(P-1). Since det(P).det(P-1)=det(PP-1)=det(I)=1, we have det(tI-B) = det(tI-A).