**Proposition 1**Prove is irrational.

Here is a proof using a traditional method (See Euclid’s Elements Book X which incorporates Theatetus work on incommensurable numbers. It includes a proof that is irrational (Proposition 22), and ends with a proof that there are infinitely many distinct irrational numbers (Proposition 115):

**Proof:**(Contradiction) Suppose is rational. Then , , and . Squaring both sides, we get or . Note, this means is even. If were odd, then for ,

So, an odd number squared is odd, thus is even. Subsitituting into the original equation we get,

Note, is even, which means is even. So, hence a contradiction. Thus, is irrational.

We also can prove this by finding the zeros (i.e., roots) of integral polynomials (i.e., a polynomial with only integer coefficients). Let us consider the integral polynomial , and look for the rational zeros (i.e., zeros which are rational number) of this polynomial. At the first glance, there seems to be infinitely many possibilities for the rational zeros. However, there is a very useful theorem for this problem, called the Rational Zero Theorem. It helps us narrow down the possible rational zeros of an integral polynomial to a list of finite rational numbers.

**Theorem 1**(The Rational Zero Theorem). Let with . If and are nonzero integers such that and (i.e., there is no common factor among and except ), then and .

**Proof:**Let such that . Assume that and So,

So,

Thus, . Since, , we have . Similary,

So, . Since, , we have .

**Remark 1**If with and (with nonzero integers and ; this means that we cannot reduce the expression further) is a zero of , then the denominator must divide the leading coefficient of and the numerator must divide the constant term of .

**Example 1**Consider the integral polynomial , Since , is not a zero of . We can also deduce the fact that is not a zero of by noting that the constant term of (which is ) is not zero. If (with and ) is a zero of , then the numerator divides the constant term (i.e., ) and the denominator divides the leading coefficient , (i.e., ). So, the possible are and . Similarly, the possible are and . Thus, the possible rational zeros are , , , and .

A special case of the Rational Zero Theorem shows us that the rational zeros of an integral polynomial with the leading coefficient 1 (i.e., a monic integral polynomial) are all integers.

**Corollary 1**Let . If (i.e., the leading coefficient is ) and the integral polynomial has a rational zero, then the rational zeros of are all integers, and they must divide the constant term .

**Proof:**Let . Assume that and has a rational zero. Let and be nonzero integers such that and is a zero of . According to the Rational Zero Theorem, the integer must divide the leading coefficient . So, must be either or . Thus, the zero of is either the integer or . Again, by the Rational Zero Theorem, we have ; so, and .

Now we are ready to prove is irrational using the Rational Zero Theorem:

**Proof:**Let us show that is an irrational number. Note that it is a zero of the polynomial . Now, possible rational zeros of this polynomial are and . Now, , , and . So, there is no rational zero for the polynomial . So, can not be a rational number, i.e., is an irrational number.

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December 6, 2011 at 8:41 am

Philip CoffeyTakes me back to the Math 300 days. Makes me happy I can still follow along with what’s being said!