# 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 â‰  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 âˆšm is irrational.

âˆš2 ,âˆš3, âˆš5, âˆš6, âˆš7, âˆš8, âˆš10,….. etc., all are irrational.

2) If m is a positive integer which is not a perfect cube , then 3âˆšm is irrational.

3âˆš2,  3âˆš3,  3âˆš4,…..  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 â‰  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 â‰  0. Then, x is a nonterminating repeating decimal, if q â‰  (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: âˆš2 is irrational

Let âˆš2 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 â‰  0.

Now âˆš2 = p/q

â‡’ 2 = p2/q2    (on squaring both sides)

â‡’ 2q2 = p2    ……..(i)

â‡’ 2 divides p2   (2 divides 2q2

â‡’ 2 divides p   (2 is prime and divides p2 â‡’ 2 divides p)

Let p = 2r for some integer r.

putting p = 2r in eqn (i)

2q2 = 4r2

â‡’ q2= 2r2

â‡’ 2 divides q2   (2 divides 2r2 )

â‡’ 2 divides p    (2 is prime and divides q2 â‡’ 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 âˆš2 is rational. So, âˆš2 is irrational.

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

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

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

n â‰ 0.

Then, âˆšp = m/n

â‡’ p = m2/n2         [on squaring both sides]

â‡’ pn2 = m2 ……..(i)

â‡’ p divides m2  (p divides pn2)

â‡’ p divides m    (p is prime and p divides m2 â‡’ p divides m)

Let m = pq for some integer q.

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

pn2 = p2q

â‡’ n2 = pq2

â‡’ p divides n2 [ p divides pq2

â‡’ 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 + âˆš3 is irrational.

If possible, let (2 + âˆš3) be rational. Then, (2 + âˆš3) is rational, 2 is rational

â‡’  {( 2 + âˆš3) – 2} is rational          [difference of rationales is rational]

â‡’ âˆš3 is rational. This contradicts the fact that âˆš3 is irrational.

The contradiction arises by assuming that (2 + âˆš3) is irrational.

Hence, (2 + âˆš3) is irrational.

### Proof 4: âˆš2 + âˆš3 is irrational.

Let us suppose that (âˆš2 + âˆš3 ) is rational.

Let (âˆš2 + âˆš3) = a, where a is rational.

Then, âˆš2 = (a – âˆš3 )    ………….(i)

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

2 = a2 + 3 – 2aâˆš3 â‡’  2aâˆš3 = a2 + 1

Hence, âˆš3 = (aÂ² +1)/2a

This is impossible, as the right hand side is rational, while âˆš3  is irrational. This is a contradiction.

Since the contradiction arises by assuming that (âˆš2 + âˆš3) is rational, hence (âˆš2 + âˆš3) 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 â‰  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 â‰  (2m x 5n).

### Examples

(i) 33/50

Now, 50 = (2×52)  and 2 and 5  is not a factor of 33.

So, 33/ 50 is in its simplest form.

Also, 50 = (2×52) = (2m Ã— 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 Ã— 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 â‰  (2m Ã— 5n) .

53 /343 is a non-terminating repeating decimal.

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