Math Proofs

Size of the power set

T. — Sat, 28 Feb 2009 02:53:01 +0000

Introduction. Let be any arbitrary set. We say a set is a subset of (written ) if every element of is also an element of . For example, , every set is a subset of itself, and the empty set is the only subset of itself.

Again consider set . We define the power set of , , as

Thus the power set of is “the set of all subsets of ”. Here are a few examples of power sets:

Let . Then the power set of is ;
Let . Then the power set of is ;
The power set of is . Notice that these two are different: the former is a set with no elements, while the latter is a set with a single element which happens to be the empty set.

We now pose an interesting question: given a set with elements (written ), what is the number of elements of its power set (written )? Peeking at concrete examples like those above, looks like the correct answer — even for the last one, since . But is it really so?

Theorem. Let denote an arbitrary set such that . Then .

Proof. The proof is by induction on the number of elements of .

For the base case, suppose . Clearly, . But the empty set is the only subset of itself, so .

Now the induction step. Suppose ; by the induction hypothesis, we know that . Let be a set with elements, namely . There are two kinds of subsets of : those that include and those that don’t. The first are exactly the subsets of , and there are of them. The latter are sets of the form , where ; since there are possible choices for , there must be exactly subsets of of which is an element. Therefore .

Quadratic formula

T. — Sat, 28 Feb 2009 01:53:13 +0000

Introduction. Let , where . We call the equation a quadratic equation. It is a well-known fact that, given and , there are at most two distinct values of satisfying the equation — called solutions or roots. For example:

The equation has one solution, namely ;
The equation has two solutions, namely and ;
The equation has no solutions. This can be shown pictorially by drawing the graph of , which never intersects the -axis.

Given a quadratic equation, its roots can be found by applying the quadratic formula:

We now intend to prove the correctness of this formula.

Theorem. Let , where . Then the solutions to the equation , if they exist, are given by the formula .

Proof. We use a proof technique called completing the square, which makes use of the binomial identity

Specifically, we want to “complete the square” by identifying it with the first two terms of the right hand side of the binomial identity. This can be readily done by setting

and solving for and , giving

Thus, the third and missing term of the square is . We now add this to both sides of the original equation, and move to the right hand side:

Now we apply the binomial identity to the left hand side of the above equation, yielding

and we take the square root of both sides, which gives

It is now clear why we went through the trouble of completing the square: the above equation is easily solved for . First divide everything by :

Now factor out of the square root:

And finally, isolate :

This is the quadratic formula, as desired.

Remark. Although we derived the quadratic formula for real-valued quadratic equations, the result holds as well for complex-valued quadratic equations; it can also be used with quadratic equations whose coefficients are themselves complex. This is because all the algebraic properties used in the proof are still valid in the field of complex numbers.

More precisely: given , , the quadratic formula can still be used to compute the roots of , where ranges over the complex numbers. In fact, the Fundamental Theorem of Algebra ensures that every quadratic equation has two complex roots — which is not true when we only consider real roots.

Irrationality of √2

T. — Fri, 27 Feb 2009 01:31:25 +0000

Introduction. Let be the set of real numbers. Intuitively, these are all the numbers that can turn up when we measure the distance between two points. The set includes the naturals , the integers , and the rationals . The latter are all the numbers that can be written as fractions; more precisely, if and , then . Every fraction can be made irreducible, i.e., we can write any rational number as where and are coprime — they have no common divisors apart from .

Clearly, all the sets mentioned so far are successively contained in each other: . But is ? The answer is negative. The remaining elements of — i.e., those not in — are aptly called irrational.

Consider a real . We say is a square root of if . It can be shown — but we will not show here — that every non-negative real number has a real-valued square root; in fact, every positive real number has two distinct square roots (one positive and the other negative), and zero is the unique square root of itself. Whenever has a square root, we denote by the non-negative square root of .

Since is a positive real number, it follows that exists in . Here we will show that is irrational. The importance of this theorem lies in the fact that it proves irrational numbers do indeed exist.

Theorem. is irrational.

Proof. Suppose is rational. Then we can find two coprime integers and , both nonzero, such that . Let us square both sides of the equation; we get . Because , we can write . Therefore is even and must be even as well. Similarly, we can write , concluding that is even. But then and are both divisible by two, contradicting the assumption that they are coprime. Therefore it must be false that is rational.

Fundamental theorem of arithmetic

T. — Fri, 27 Feb 2009 00:28:57 +0000

Introduction. Let be the set of natural numbers. We say is a prime number if has exactly two distinct divisors: and itself. The first few prime numbers are:

Given , a prime factorization of is a product of naturals such that the are all prime. Notice that a prime factorization may contain multiple occurrences of the same prime number, but the order in which the appear is not important. For example, and are prime factorizations of and , respectively. Clearly, if is a prime number, itself is its unique prime factorization. Also, the number is special in that its unique prime factorization is the empty product.

Considering an arbitrary natural number , two interesting questions arise. First, is guaranteed to have a prime factorization? Second, if it does exist, is it unique (up to rearrangement)? The Fundamental Theorem of Arithmetic answers affirmatively to these two questions.

Theorem. Every natural number has a unique prime factorization.

Proof. The proof is divided in two parts. First we prove the existence of a prime factorization for every natural number, and then we prove its uniqueness.

We prove existence first. Let be the set of all natural numbers without a prime factorization. Aiming for a contradiction, assume . Clearly, has a minimal element; call it . Notice that cannot be prime, for otherwise it would be a prime factorization of itself. Therefore we can write , where and are natural numbers greater than . This implies . Because was chosen to be the smallest number in , and must be outside . Therefore we can find a prime factorization of by concatenating prime factorizations of and , which are guaranteed to exist. This contradicts the hypothesis that , concluding the proof of existence.

Now we prove uniqueness. Let be a natural number admitting two distinct prime factorizations, namely and . We will show that the two factorizations must in fact be equal (up to rearrangement). Consider Because divides it must also divide one of the ; without loss of generality, assume divides . Because and are both prime, it follows that . Now we can cancel and out of their respective sequences and proceed by induction until one of the sequences, say the first one, has been reduced to an empty product. We have thus proved that for . It only remains to be proved that . To that effect notice that, if n' title='m > n' class='latex' />, the reduction implies , which is impossible unless the factors in the right side are all .