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Previous: Induction in Calculus
Irrational Numbers All of the real numbers that we deal with in everyday life can be categorised in one of two ways. They are either “rational numbers”, or they are “irrational numbers”. There are no other options. The rational numbers can be expressed as the ratio of two integers. Thus, for example, the rational number n can be expressed as
where a and b are integers, b is not equal to zero, and a and b have no common factors. If they did, the common factor could be removed by division. For example, in the ratio 6/4, 6 and 4 have the common factor 2, which can be divided out to give the equivalent ratio 3/2. Note that since b can be equal to 1, all the integers are rational numbers. Irrational numbers cannot be expressed as a fraction in this way. One consequence of this is that irrational numbers cannot be represented as a decimal of finite length or a decimal that shows a repeating pattern of numbers. For example: n = 1.456342 is a decimal of finite length, and hence is rational (fractional form: 1456342/1000000). Similarly, the number n = 0.142857142857 … is a decimal of infinite length, but it has a repeating sequence (142857) and so is rational (in fact this number is just 1/7). On the other hand, the number π (the ratio of the circumference to the diameter of a circle) is an irrational number. The first few decimal places of π are: π = 3.141592653 … The decimal expansion of π is neverending, and will show no repetition of digits of the type that we saw above.
Square Root of 2 One of the bestknown of the irrational numbers (apart from π) is the square root of 2, written as √2. If you imagine a square with each side having a length of 1 unit, then the length of a diagonal of the square is √2 units. That √2 is irrational can be proved by a number of means, including some clever geometrical arguments. However, the proof by contradiction is perhaps the bestknown of the proofs, and is the one we will consider here. As with all proofs by contradiction, we begin by writing the statement we wish to prove: S = The number √2 is irrational. The converse of the statement S is easy to write down, since real numbers are either rational or irrational: T = The number √2 is rational. To prove the correctness of S, we will assume that statement T is true, and we will see if this leads to some kind of contradiction or other nonsensical result. Let’s assume that T is true. In this case, noting what we said in the previous section, we can express √2 in the following form:
where a and b are integers with no common factors, as required by our definition of a rational number, as discussed above. Now, the square of an even number is even, and the square of an odd number is odd. (This is easy enough to demonstrate – let n be an integer. The square of the even number 2n is 4n^{2}, which is clearly even, and the square of the odd number 2n+1 is 4n^{2}+4n+1, which is clearly odd). With this in mind, we note that a must be even and hence we can write
where c is another integer. Now, substituting this into the previous expression gives
This shows that b is also an even number, and hence
where d is yet another integer. This is all every interesting. At the beginning, we assumed that the integers a and b have no common factors, and yet we have just shown that if √2 is rational, then both a and b would be even, and hence have a common factor of 2. This contradicts our original assertion of no common factors, and demonstrates that the statement T is false. Hence, statement S is true, and √2 is an irrational number.
Square Root of 3 In this section we will consider the square root of 3, √3. The proof that this is irrational requires a little more thought than the proof that √2 is irrational, but proceeds in broadly the same manner. The statement we need to prove can be written as S = The square root of 3 is irrational. The converse of this statement is as follows: T = The square root of 3 is rational. If statement T is true, then our original assertion in statement S is false. So, let us assume that statement T is true, and see if we can find a contradiction. If √3 is rational, then we can define integers a and b such that
where a and b are integers with no common factors. Now, if b is an odd number, then a is also an odd number, since the number 3 is odd. Similarly, if b is even, then a is also even. However, the case of a and b even means that they have at least one common factor of 2. Thus, in this case, a and b would not be in their lowest forms which contradicts our original specification of the numbers a and b. We therefore are forced to assume that the integers a and b are both odd. If a and b are both odd, then they can be written in the form
where m and n are integers. Substituting into the expression above gives
We have found our contradiction, as we cannot find integers m and n that satisfy this expression. Why not? Clearly, the left hand side of this expression is even. The first two terms on the right are even, but the number 1 is odd, and so the righthand side is itself odd. This is clearly an impossibility, and so the statement T is false. Hence, statement S is true, and √3 is an irrational number.
