We have studied Rational Numbers and their properties. Every rational number is of the form p/q, where p is an integer and q is a natural number; like 3/5, 1/3, -7/11. We also know how to add two rational numbers and how to multiply two rational numbers.
Let us revise, click here →Real numbers exercise 1.2.1
Now let us study real numbers,
Real numbers are those which comprises of both rational numbers and irrational numbers.
The expansion of a rational number, after carrying out the division, consists of:
(1) integer part (the integer which you get before the decimal point);
(2)some finite number of digits which is not a repeating part and which occur soon after the decimal point;
(3) a string of digit which repeats itself and which start repeating after the non-repeating part.
|number||expansion||integer part||non-repeating(impure) part||repeating(periodic) part||length|
The repeating part of a rational number is called the periodic part or simply the period of the rational number. The non-repeating or nonrecurring part is some times called the impure part of the rational number. Thus a rational number has three parts: integer part: impure part: periodic part. The number of digits in the period is called its length.
There may not be any impure part for a rational number: like 1/3 = 0.3333333… in such case we say the rational number is purely periodic. If the periodic part starts after the non-repeating part, we say the expansion is eventually periodic. Thus 0.333333… is denoted by
Theorem 1: Any eventually periodic decimal number corresponds to a rational number.
Let us take an eventually periodic decimal expansion. It is of the form a.p1.p2…pk.
q1.q2…..ql ,where a is an integer, p1.p2…pk is the non-repeating part and q1.q2…..ql is the periodic part. We consider only part r = 0. p1.p2…pk q1.q2…..ql and show that this is a rational number. The given number is simply a + r. We multiply r by 10k and obtain
10k.r = p1.p2…pk
q1.q2…..ql ———————– (1)
Again multiplying this by 10k, we obtain
10l(10k.r) = p1.p2…pkq1q2…
ql . q1.q2…..ql ———————– (2)
Observe b = p1.p2…pk and c = p1.p2…pkq1q2…ql are two numbers in the base 10.
Subtracting (2) from (1), we obtain,
(10k+l – 10k) r = c – b
Hence, r = (c – b)/( (10k+l – 10k) r.
This shows that r is a rational number. Finally the given number is a+r is also rational since a is an integer and r is rational number.
A number is rational if and only if it has a decimal expansion which is either terminating or eventually periodic.
A number is rational if and only if it has a decimal expansion which is eventually periodic.
click here Real Numbers exercise 1.2.2 for solved problems.
1.2.3 Irrational numbers
A number having an infinite decimal expansion which is not eventually periodic is called an irrational number. Some examples for irrational numbers: √2, √3, pi, 7√5, In fact, √p is irrational for any prime number p.
For any two rational numbers r1 and r2 such that r1 < r2, we can find an irrational number i1 such that r1 <i1 < r2 . in fact we can find infinitely many rational numbers between any two rational numbers. This property is true for any two distinct rational numbers.
Let us find 3 rational numbers between ½ and 2/3
½ = 0.5 = 0.4999999…, and
2/3 = 0.666666…
Consider the number a = 0.50500500050000… This is larger than 0.5 and is irrational. But this is smaller than 0.6666…, Similarly, the number b= 0.60600600060000…is an irrational number and is smaller than 0.66666… Consider another number c = 0.56566566656666… This is an irrational and lies between a and b. (we can write it as a < c <b)
For any two irrational numbers i1 and i2 such that i1 < i2 , we can find a rational number r1 such that i1 < r1 < i2. In fact we can find infinitely many rational numbers between any two irrational numbers. This property is true for any two distinct irrational numbers.
Let us find 3 rational numbers between 1.0100100010000… and 1.012303003000…
Observe these two are irrational numbers. Consider the numbers r1 = 1.010010001000…, r2 = 1.011222222…, r3 = 1.0122222222… These are three rational numbers and
1.0100100010000… < r1 < r2 < r3 < 1.012303003000…
For related solved problems click on Real Numbers exercise 1.2.3
1.2.4 Real Number System
The collection of all rational numbers and irrational numbers together is the set of all Real numbers. This is denoted by ℜ. Thus every real number is either a rational number or an irrational number. We can put two binary operations on ℜ which are called addition (denoted by +) and multiplication (denoted by . or x). With these operations ℜ has all the nice properties we have seen in Q, the set of all rational numbers. With the background we have, it is not possible to properly define these operations and prove the properties of these operations. We take them for granted and record all these properties:
(1) closure property: The set ℜ of real numbers is closed under addition and multiplication. Thus we can add any two real numbers and we get another real number; for example 2 and √2 can be added to get (2 + 2√2) which again is a real number; for example (√5 x ∛3) is a real number. More generally, for any two real numbers a and b, both a + b and a x b are real numbers.
(2) Commuutative property: Just like rational nummbers, addition and multiplication on real numbers also have commutative
For example we have,
∛7 + 1.234 = 1.234 + ∛7 ; √2 x ∛1.75 = ∛1.75 x √2.
(3) Associative Property: Suppose you have 3 real numbers, say a, b and c, then you have
(a + b) + c = (a + b) + c;
(a x b) x c = a x (b x c)
Thus, for exampple,
(2 + √3) + π = 2 + (√3 + π);
(0.1234567891011… x 1.8) x √5 = 0.1234567891011… x (1.8 x √5)
(4) Distributive Property: In real number system, the multiplication is distributive over addition. If a, b and c are three real number, then
a x (b + c) = (a x b) + (a x c);
π x (√17 + 1/17) = (π x √17) +(π x 1/17)
(5) Additive and Multiplicative Identities: In real number system, the multiplication is distributive over addition, such that,
a + 0 = a, for all real numbers a.
a x 1 = a, for all real numbers a.
Thus we see that, for example, (√(1+5√2)) + 0 = (√(1+5√2)) and (π x √8) x 1 = (π x √8). We say that 0 is the additive identity and 1 is the multipilcative identity.
(6) Additive inverse and multiplicative inverse: For every real number a, there exists a unique real number denoted by (-a) such that, a + (-a) = 0 = (-a) + a. We say that (-a) is the additive inverse of a. Thus for example the additive inverse of (1 + √5) is -(1 + √5) since, (1 + √5) + [-(1 + √5)] = 0.
Similarly, if b is anon-zero real number then there exists a unique real number denoted by b-1 such that b x (b-1) = 1 = (b-1) x b. This is called multiplicative inverse of b. We also denote b-1 as 1/b. The most important thing here is that the multiplicative of a real number b exists if and only if b ≠ 0. Thus if we take the real number √13, then its multiplicative inverse 1/√13 exists as real number and √13 x (1/√13) = 1.
(7) Cancellation laws: Suppose a,b and c are real numbers such thta a x c = b x c. If c ≠ 0, then a = b. This means we can cancel non-zero quantity on both the sides of an equality. This is a very useful proprety while solving equations. Similarly, if a + c = b + c, then cancelling c on both the sides we get a = b. This is again useful in solving equations, since x + a = b gives x = b – a.
(8) Ordering property: Given any two real numbers a and b, we can compare the,: either a = b or a < b or b < a, and exactly one of these three relation holds. Thus we can order the real numbers.
In practice, comparing two real numbmers may be harder. The rational numbers and irrational numbers are insepereably interwind in this system: between any two real numbers, we can find infinitely many rational and irrational numbers.
Suppose a and b are real numbers such that a < b and let c be a positive real number, i.e., c > 0. Then a x c < b x c. Thus we can multiply an inequality by a positive real number without affecting it. If d > 0, then a x d > b x d; thus the inequality changes when it is multiplied by a negative number. For example: we have 7 < 10. If we multiply both sides by -6, 7 x (-6) = -42 and 10 x (-6) = -60. We know -42 > -60. Thus 7 x (-60) > 10 x (-6), and the inequality is reversed. Hence iit is very important to check the sign og the number before we multiply an inequality by it.
Here we define subtraction and division using addition and multiplication respectively. Suppose we want to subtract a real number b by another real number a. We define this by a + (-b), where -b is the additive inverse of b and we denote this as a – b. Similarly, if b ≠ 0 is a real number and a is some other real number, we define the division of a by b to be a x (b-1) where (b-1) is multiplicative inverse of b. We denote this by a/b.
We know that Q contains Z, the set of all integers. Now ℜ contains Q and irrational numbers. We can put all these pictorically.
Thus, we see that,
N ⊂ W ⊂ Z ⊂ Q ⊂ R, J ⊂ R and R = Q U J
We know there is no rational number p/q such that (p/q)² = 2. However, we have included √2 in the real nummber system R and (√2)² = 2.
Given any positive integer n and positive real number a, there is a unique real number b such that bn = a. If n is an odd positive integer, then for any real number a there is a unique real number b such that bn = a.
Example: For any real number a, prove that a x 0 = 0.
For any real number a, by the additiveidentity property of 0. Using distributive property
a x 0 = a x (0 + 0) = a x 0 + a x 0.
We can cancel a x 0 on both sides and get a x 0 = 0.
Example: Prove that (-1) x (-1) = 1
Since 1 is the multiplicative identity, we know that 1 x (-1) = (-1). Using distributive property,
0 = (1 – 1) x (-1) = 1 x (-1) + (-1) x (-1)
Thus, (-1) x (-1) + (-1) = 0. Adding 1 on both sides, we get,
(-1) x (-1) = 1
Example : Prove that the square of a real number is always non-negative.
Suppose a is a real number. Then either a > 0 or a = 0 or a < 0 and only one of these is true.
Suppose a > 0. Then a x a > 0 x a = 0, since we are multiplying by a positive real number. Hence a² > 0.
If a = 0. then a² > 0.
Suppose a < 0. Then a x a < 0 x a, as the inequality changes when multiplied by a negative number. Again we obtain a² > 0.
We conclude that a² > 0 when a is either positive or negative; and a² = 0 when a = 0. Thus a² ≥ 0 for any real number a.
For related solved problems click on Real numbers exercise 1.2.4
Representation Of Irrational numbers on Number Line
We can use number line to represent irrational nummbers as we represent rational numbers on number line.
One of the classical result in geometry dating back to Greek period is Pthagoras’ theorem. Given a right angle triangle ABC with ∠B = 90°,this theorem states that AB² + BC² = CA². This is very useful in constructing newer lengths as we see now.
Take an infinite line and select a point on it which is aclled orgin. This is the point used to represent 0. Now let us take a unit length on it, say OA = 1. Draw a perpenndicular AK to the line at K such that AK = 1 Join OK. Pythagoras’ theorem says that OA² + AK² = OK² Thus OK² = 1² + 1² = 1 + 1 = 2. We obtain OK = √2. With O as centre and OK as radius, draw an arc cutting the line at B. Then OB = OK = √2. Thus √2 is represented by the point B on the line.
Suppose we want to represent √3 on the number line. We draw BL perpendicular to the line at B such that BL = 1. Join OL. Using Pythagoras theorem, we get OL² = OB² + BL². But OB = √2 so that OB = 2. Hence OL² = 3. and this gives OL = √3. Now draw an arc with O as centre and OL as radius to cut the line in C. Then we get OC = √3.
We can continue this process in steps; knowing 2, we can represent √5, since 5 = √(2²+1²); knowing √5, we can represent √6 and so on. More generally, suppose we have represented √n on the number line by apoint N. we can use this information to represent √(n+1) on the nummber line.
Draw NS perpendicular to the line at N such that NS = 1. Join OS. The pythagoras’ theorem gives OS² = ON² + NS² = (n + 1). Hence OS = √(n+1). With O as the centre and OS as radius, draw an arc cutting the line at R. Then OR = OS = √(n+1). Hence R represents √(n+1) on the number line.
Example: Represent √7 on the number line.
Let O be the origin on the line l. Let A be on the line such that OA = 1. Draw AB perpendicular to OA at A such that AB = 1. Then OB² = OA² + AB² = 1² + 1² = 2. Thus OB = √2. With O as centre and OB as radius draw an arc cutting the line at C. Then OC = OB = √2. Again draw CD perpendicular to l such that CD = 1. Then OD² = OC² + CD² = 2 + 1 = 3. Thus OD = √3. Draw an arc with O as centre and OD as radius to cut l in E. Then OE = OD = √3. Draw FE perpendicular to l at E such that EF = 2 and join OF. Now,
OF² = OE² + EF² = 3 + 4 = 7.
Hence OF = √7. With O as centre and OF as radius, cut l at G. Then OG = OF = √7. Thus G represnts √7 on the line l.
For related solved problems click Real Numbers exercise 1.2.5