# 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 let’s 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 ≠ 0 are called rational numbers.

**Examples:**

1)All natural numbers are rational,

1, 2, 3, 4, 5…….. 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 √m is irrational.

√2 ,√3, √5, √6, √7, √8, √10,….. etc., all are irrational.

2)If m is a positive integer which is not a perfect cube , then^{3}√m is irrational.

^{3}√2,^{3}√3,^{3}√4,….. etc., all are irrational.

3)Every Non Repeating and Non Terminating Decimals are Irrational Numbers.

0.1010010001…… is an non-terminating and non repeating decimal. So it is irrational number.

0.232232223…….. is irrational.

0.13113111311113…… 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 ≠ 0. Then, x is a terminating decimal only when q is of the form (2^{m}x 5^{n}) 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 ≠ 0. Then, x is a nonterminating repeating decimal, if q ≠ (2^{m}x 5^{n}).**Theorem 2:**Let x be a rational number whose simplest form is p/q, where p and q are integers and q = 2^{m}x 5^{n}then x has a decimal expansion which terminates.

**Proof 1: √2 is irrational**

Let √2 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 ≠ 0.

Now √2 = p/q

⇒ 2 = p

^{2}/q^{2 }(on squaring both sides)⇒ 2q

^{2 }= p^{2}……..(i)⇒ 2 divides p

^{2 }(2 divides 2q^{2 })⇒ 2 divides p (2 is prime and divides p

^{2}⇒ 2 divides p)Let p = 2r for some integer r.

putting p = 2r in eqn (i)

2q

^{2 }= 4r^{2}⇒ q

^{2}= 2r^{2}⇒ 2 divides q

^{2}(2 divides 2r^{2})⇒ 2 divides p (2 is prime and divides q

^{2}⇒ 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 √2 is rational. So, √2 is irrational.

### **Proof 2: Square roots of prime numbers are irrational **

Let p be a prime number and if possible, let √p be rational.

Let its simplest form be √p=m/n, where m and n are integers having n no common factor other than 1, and

n ≠0.

Then, √p = m/n

⇒ p = m

^{2}/n^{2 }[on squaring both sides]⇒ pn

^{2 }= m^{2}……..(i)⇒ p divides m

^{2}(p divides pn^{2})⇒ p divides m (p is prime and p divides m

^{2}⇒ p divides m)Let m = pq for some integer q.

Putting m = pq in eqn (i), we get:

pn

^{2}= p^{2}q^{2 }⇒ n

^{2 }= pq^{2}⇒ p divides n

^{2}[ p divides pq^{2}]⇒ p divides n [p is prime and p divides n

^{2}= 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 + √3 is irrational.**

If possible, let (2 + √3) be rational. Then, (2 + √3) is rational, 2 is rational

⇒ {( 2 + √3) – 2} is rational [difference of rationales is rational]

⇒ √3 is rational. This contradicts the fact that √3 is irrational.

The contradiction arises by assuming that (2 + √3) is irrational.

Hence, (2 + √3) is irrational.

**Proof 4: √2 + √3 is irrational.**

Let us suppose that (√2 + √3 ) is rational.

Let (√2 + √3) = a, where a is rational.

Then, √2 = (a – √3 ) ………….(i)

On squaring both sides of (i), we get:

2 = a^{2}+ 3 – 2a√3 ⇒ 2a√3 = a^{2}+ 1

Hence, √3 = (a² +1)/2a

This is impossible, as the right hand side is rational, while √3 is irrational. This is a contradiction.

Since the contradiction arises by assuming that (√2 + √3) is rational, hence (√2 + √3) 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 ≠ 0. Then,

**(i) x is a terminating decimal only when q is of the form (2 ^{m} x 5^{n}) for some non-negative integers m and n. **

**(ii) x is a nonterminating repeating decimal if q ≠ (2 ^{m} x 5^{n}).**

**Examples**

(i) 33/50Now, 50 = (2×5

^{2}) and 2 and 5 is not a factor of 33.So, 33/ 50 is in its simplest form.

Also, 50 = (2×5

^{2}) = (2^{m}× 5^{n}) where m = 1 and n = 2.53/343 is a terminating decimal.

33/50 = 0.66.

(ii) 41/1000Now, 1000 = (2

^{3}x5^{3}) = and 2 and 5 is not a factor of 41.So, 41/ 1000 is in its simplest form.

Also, 1000 = (2

^{3}x2^{3}) = (2^{m}× 5^{n}) where m = 3 and n = 3.4 /1000 is a terminating decimal.

41/1000 = 0.041

(iii) 53/343Now, 343 = 7

^{3}and 7 is not a factor of 53.So, 53/ 343 is in its simplest form.

Also, 343 =7

^{3}≠ (2^{m}× 5^{n}) .53 /343 is a non-terminating repeating decimal.

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