Decimal Expansions of Rational Numbers
Real numbers are simply the combination of rational and irrational numbers, in the number system. In general, all the arithmetic operations can be performed on these numbers and they can be represented in the number line, also. So in this article letβs discuss some rational and irrational numbers and their proof.
Rational Numbers
A number of the form p/q, where p and q are integers and q β 0 are called rational numbers.
Examples:
1) All natural numbers are rational,
1, 2, 3, 4, 5β¦β¦.. all are rational numbers.
2) Whole numbers are rational.
0,1, 2, 3, 4, 5, 6,,,,,, all are rational.
3) All integers are rational numbers.
-4.-3,-2,-1, 0, 1, 2, 3, 4, 5,,,,,,,, all are rational numbers.
Irrational Numbers
The numbers which when expressed in decimal form are expressible as non-terminating and non-repeating decimals are known as irrational numbers.
Examples:
1) If m is a positive integer which is not a perfect square, then βm is irrational.
β2 ,β3, β5, β6, β7, β8, β10,β¦.. etc., all are irrational.
2) If m is a positive integer which is not a perfect cube , then 3βm is irrational.
3β2, 3β3, 3β4,β¦.. etc., all are irrational.
3) Every Non Repeating and Non Terminating Decimals are Irrational Numbers.
0.1010010001β¦β¦ is an non-terminating and non repeating decimal. So it is irrational number.
0.232232223β¦β¦.. is irrational.
0.13113111311113β¦β¦ is irrational.
Nature of the Decimal Expansions of Rational Numbers
- Theorem 1: Let x be a rational number whose simplest form is p/q, where p and q are integers and q β 0. Then, x is a terminating decimal only when q is of the form (2m x 5n) for some non-negative integers m and n.
- Theorem 2: Let x be a rational number whose simplest form is p/q, where p and q are integers and q β 0. Then, x is a nonterminating repeating decimal, if q β (2m x 5n).
- Theorem 2: Let x be a rational number whose simplest form is p/q, where p and q are integers and q = 2m x 5n then x has a decimal expansion which terminates.
Proof 1: β2 is irrational
Let β2 be a rational number and let its simplest form is p/q.
Then, p and q are integers having no common factor other than 1, and q β 0.
Now β2 = p/q
β 2 = p2/q2 (on squaring both sides)
β 2q2 = p2 β¦β¦..(i)
β 2 divides p2 (2 divides 2q2 )
β 2 divides p (2 is prime and divides p2 β 2 divides p)
Let p = 2r for some integer r.
putting p = 2r in eqn (i)
2q2 = 4r2
β q2= 2r2
β 2 divides q2 (2 divides 2r2 )
β 2 divides p (2 is prime and divides q2 β 2 divides q)
Thus 2 is common factor of p and q. But, this contradict the fact that a and b have common factor other than 1. The contradiction arises by assuming that β2 is rational. So, β2 is irrational.
Proof 2: Square roots of prime numbers are irrational
Let p be a prime number and if possible, let βp be rational.
Let its simplest form be βp=m/n, where m and n are integers having n no common factor other than 1, and
n β 0.
Then, βp = m/n
β p = m2/n2 [on squaring both sides]
β pn2 = m2 β¦β¦..(i)
β p divides m2 (p divides pn2)
β p divides m (p is prime and p divides m2 β p divides m)
Let m = pq for some integer q.
Putting m = pq in eqn (i), we get:
pn2 = p2q2
β n2 = pq2
β p divides n2 [ p divides pq2]
β p divides n [p is prime and p divides n2 = p divides n].
Thus, p is a common factor of m and n. But, this contradicts the fact that m and n have no common factor other than 1. The contradiction arises by assuming that /p is rational. Hence, p is irrational.
Proof 3: 2 + β3 is irrational.
If possible, let (2 + β3) be rational. Then, (2 + β3) is rational, 2 is rational
β {( 2 + β3) β 2} is rational [difference of rationales is rational]
β β3 is rational. This contradicts the fact that β3 is irrational.
The contradiction arises by assuming that (2 + β3) is irrational.
Hence, (2 + β3) is irrational.
Proof 4: β2 + β3 is irrational.
Let us suppose that (β2 + β3 ) is rational.
Let (β2 + β3) = a, where a is rational.
Then, β2 = (a β β3 ) β¦β¦β¦β¦.(i)
On squaring both sides of (i), we get:
2 = a2 + 3 β 2aβ3 β 2aβ3 = a2 + 1
Hence, β3 = (aΒ² +1)/2a
This is impossible, as the right hand side is rational, while β3 is irrational. This is a contradiction.
Since the contradiction arises by assuming that (β2 + β3) is rational, hence (β2 + β3) is irrational.
Identifying Terminating Decimals
To Check Whether a Given Rational Number is a Terminating or Repeating Decimal Let x be a rational number whose simplest form is p/q, where p and q are integers and q β 0. Then,
(i) x is a terminating decimal only when q is of the form (2m x 5n) for some non-negative integers m and n.
(ii) x is a nonterminating repeating decimal if q β (2m x 5n).
Examples
(i) 33/50
Now, 50 = (2Γ52) and 2 and 5 is not a factor of 33.
So, 33/ 50 is in its simplest form.
Also, 50 = (2Γ52) = (2m Γ 5n) where m = 1 and n = 2.
53/343 is a terminating decimal.
33/50 = 0.66.
(ii) 41/1000
Now, 1000 = (23x53) = and 2 and 5 is not a factor of 41.
So, 41/ 1000 is in its simplest form.
Also, 1000 = (23x23) = (2m Γ 5n) where m = 3 and n = 3.
4 /1000 is a terminating decimal.
41/1000 = 0.041
(iii) 53/343
Now, 343 = 73 and 7 is not a factor of 53.
So, 53/ 343 is in its simplest form.
Also, 343 =73 β (2m Γ 5n) .
53 /343 is a non-terminating repeating decimal.
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