Properties of Rational and Irrational Numbers

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Abner Ward
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1 Properties of Rational and Irrational Numbers September 8, 2016 Definition: The natural numbers are the set of numbers N = {1, 2, 3,...}, and the integers are the set of numbers Z = {..., 2, 1, 0, 1, 2,...}. Note that the natural numbers are the positive integers. Natural numbers are integers, but not necessarily the other way around. Both N and Z are closed under addition and multiplication. This means that if x and y are both either natural numbers or integers, then (x + y) and (x y) are, too. Definition: A rational number is a number q that can be expressed as the ratio of an integer z to a natural number n: q = z/n. The set of all natural numbers is called Q. Definition: An irrational number x is a number that is not rational, i.e. which there is no pair (z, n) with z Z and n N such that x = z/n. for Theorem 0.1 A number with a terminating decimal representation is rational. Terminating decimal representation means the number can be written in a form like 0.123, which stops after a finite number of digits. We can see that x = 0.123, for example, is a rational number: x = x = 123 x = 123/1, 000 x can thus be expressed as an integer divided by a natural number. 1

2 Theorem 0.2 A number with a repeating decimal representation is rational. Take as an example the number x = 0.123, where the line means that the decimal repeats those three digits for ever. We can show that x is rational as follows: x = x = x = x = x = x = x 999 x = 123 x = 123/999 Theorem 0.3 An irrational number has a decimal representation that neither terminates or repeats. Proof: If x is any number that has a decimal representation that terminates or repeats, it is a rational number by the previous two theorems. Thus x cannot be irrational. Theorem 0.4 If r and R are rational numbers, then so are (r + R) and (r R). If r 0, then (1/r) is also a rational number. Proof: Since r and R are rational numbers, there must exist integers a and b and natural numbers c and d, such that r = a/c and R = b/d. Then r + R = a c + b ad + bc =. d cd Because c and d are both N, by closure so is (c d). Since a, b, c and d are all integers, by closure so is (ad + bc). Thus, r + R can be written as an integer divided by a natural number, and is rational. Similarly, we have r R = a c b d = ab cd. By the same argument as before, ab is an integer and cd is a natural number. Thus (r R) is rational, as well. Finally, if r 0 is rational, then there are z Z and n N such that r = z/n. Since r 0, z must not equal 0 either. That means that there is a natural number m such that z = +m or z = m. So r = ±m/n and (1/r) = (±n)/m, that is, an integer divided by a natural number, and is rational. 2

3 Theorem 0.5 If i is an irrational number, and r is a rational number, then (i + r) is an irrational number; and, if r 0, (i r) is an irrational number as well. Proof: We use a proof by contradiction. We start by asking ourselves, what would happen if (i + r) actually were a rational number, R, say? Then i + r = R i = R r i = R + ( 1) r But the last number is rational, by the previous theorem. Thus (i+r) can only equal a rational number if i itself is rational. So if i is irrational, (i + r) must be irrational, too. Similarly, if a rational number r 0 we suppose that (i r) is rational and show that this leads to a contradiction. In this case i r = R i = R/r i = R (1/r) But, again by the previous theorem, this means that the right-hand side, and therefore i, would be rational. Thus, if r 0 is rational and i is irrational, (i r) must be an irrational number, too. Theorem 0.6 Let D be a natural number that is not an exact square. Then D is an irrational number. Proof: 1. We have, by hypothesis, D N, and D is not an exact square, i.e., it lies between two adjacent exact squares in N. We can express this by saying that there exists some natural number n such that n 2 < D < (n + 1) Our strategy is to assume that there is some D with these properties whose square root D is a rational number. We wish to show that this leads to a contradiction. The square root function, however, is not the best form to represent this condition. We can reexpress it with an equivalent condition holding just for natural numbers multiplied and added together. 3

4 3. This condition is the following: if there is a pair of natural numbers (t, u) that satisfy the equation t 2 Du 2 = 0, then D is a rational number. To see this, note that t 2 Du 2 = 0 means that t 2 = Du 2, or D = t 2 /u 2, or D = t/u, which of course means that D is rational. 4. Unfortunately though, if the pair (t, u) is a solution, then so are all pairs (mt, mu) where m is any natural number. To get a unique natural number solution of t 2 Du 2 = 0 out of all of these solutions, we will impose one more requirement: we select the pair (t, u) with the smallest value of u. If D is rational, there is always such a pair, and it is unique. 5. Our goal now is to show that the existence of this unique pair (t, u) is inconsistent with the requirement in (1) that n 2 < D < (n + 1) 2 for some natural number n. We will do this by finding another pair of natural numbers (t, u ) solving the same equation, but for which u < u. This will show our assumption that we can pick a smallest such u is not correct. Thus D cannot be rational, and in fact must be irrational. 6. So here we go. We have t 2 Du 2 = 0 with n 2 < D < (n + 1) 2, where all the symbols stand for natural numbers. n 2 < D < (n + 1) 2 (D is a non-square natural number) n 2 t 2 < Dt 2 < (n + 1) 2 t 2 (previous line times t 2 ) n 2 t 2 < D 2 u 2 < (n + 1) 2 t 2 (since t 2 = Du 2 ) nt < Du < (n + 1)t = nt + t 0 < (Du nt) < t (by subtracting nt) (by taking the square root) 7. So if we define t = (Du nt), we can see that t N (because D, u, n, t are all natural numbers and t > 0), and also that t < t. 8. Now we do the same thing for u. n 2 < D < (n + 1) 2 (D is a non-square natural number) n 2 u 2 < Du 2 < (n + 1) 2 u 2 (previous line times u 2 ) n 2 u 2 < t 2 < (n + 1) 2 u 2 (since t 2 = Du 2 ) nu < t < (n + 1)u = nu + u 0 < (t nu) < u (by subtracting nu) (by taking the square root) 9. So if we define u = (t nu), we can see that u N, and also that u < u. 4

5 10. Now we show that t 2 Du 2 = 0. This will establish the contradiction because (t, u ) is a solution in natural numbers to our equation, but u < u is smaller than our smallest solution. t 2 Du 2 = (Du nt) 2 D(t nu) 2 (definition of t and u ) = D 2 u 2 2Dntu + n 2 t 2 Dt 2 + 2Dntu n 2 Du 2 = D(Du 2 t 2 ) n 2 (t 2 Du 2 ) (multiplying it out) = (n 2 D) (t 2 Du 2 ) (simplifying) = (n 2 D) 0 (since (t 2 Du 2 ) = 0) = Since t 2 Du 2 = 0 and u < u, our assumption that u is the smallest solution of x 2 Dy 2 = 0 is incorrect. Thus D cannot be written as the ratio of an integer divided by a natural number, and is an irrational number. Theorem 0.7 Irrational numbers are not closed under addition and multiplication. Proof: By the previous theorems, x = 2 is irrational, and so is y = 1 + ( 1) 2. But x + y = 1 and x x = 2, which are both rational numbers. This counterexample shows that the irrational numbers are not closed under addition and multiplication. 5

Classify, graph, and compare real numbers. Find and estimate square roots Identify and apply properties of real numbers.

Real Numbers and The Number Line Properties of Real Numbers Classify, graph, and compare real numbers. Find and estimate square roots Identify and apply properties of real numbers. Square root, radicand,