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How to Convert a Rational Number into a Terminating Decimal?

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  • Last Updated : 01 Feb, 2022
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Real numbers are merely the mixture of rational and irrational numerals, in the numeral system. In known, all the arithmetic functions can be achieved on these numerals and they can be expressed in the number line, also. So in this article let’s examine some rational and irrational numbers and their proof.  

Rational Numbers

A numeral of the form p/q, where p and q are integers and q ≠ 0 are called rational numbers.

Examples:

All raw numerals are rational,

1, 2, 3, 4, 5…….. all are rational numbers.

Whole numerals are rational.

0,1, 2, 3,,,,,,, all are rational.

All integers are rational numerals.

-2,-1, 0, 1, 2, 3,,,,,,,,, all are rational numbers.

Irrational Numbers

The digits that when represented in decimal format are expressible as non-terminating and non-repeating decimals are understood as irrational numbers.  

Examples:

If n is an optimistic integer that is not an ideal square, then √n is irrational.

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

If n is an optimistic integer that is not an ideal cube, then 3√n is irrational.  

2√1,  4√3,  5√4,…..  etc., all are irrational.  

Every Non-Duplicating and Non-Terminating Decimals are Irrational Numbers.

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

0.232232223…….. is irrational.

0.13113111311113…… is irrational.

Characters of the Decimal Expansions of Rational Numerals

Theorem 1: Let x be a rational numeral whose easiest 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 (2r x 5s) for some non-negative integers r and s.

Theorem 2: Let x be a rational numeral whose easiest form is p/q, where p and q are integers and q ≠ 0. Then, x is a nonterminating repeating decimal, if q ≠ (2r x 5s).

Theorem 3: Let x be a rational numeral whose easiest form is p/q, where p and q are integers and q = 2r x 5s then x has a decimal expansion which removes.

Proof 1: √3 is irrational

Solution:

Let √3 be a rational numeral and allow its easiest formation is p/q.

Then, p and q are integers holding no standard element other than 1, and q ≠ 0.

Now √3 = p/q  

⇒ p = √3/ q    (on squaring both sides)

⇒ p2 = 3q2   

p2 / 3 = q2  ……..(i)

(1) indicates that 3 is an element of p. (Since we understand that by theorem, if a is a prime numeral and if a ranges p2, then a divides p, where a is a positive integer)

Here 3 is the prime numeral that splits p², then 3 divides p and therefore 3 is an element of p.

Since 3 is a factor of p, we can write p = 3c (where c is a constant). Substituting p = 3c in (1), we get,

(3c)² / 3 = q²

9c²/3 =  q²  

3c²  =  q²  

c²  =  q² /3 ——- (2)

Thus 3 is a factor of q (from 2)

Equation 1 offers 3 as a factor of p and Equation 2 indicates that 3 is a factor of q. This is a paradox to our belief that p and q are co-peaks. So, √3 is not a rational numeral. Thus, the root of 3 is irrational.

Proof 2: Square roots of prime numbers are irrational  

Solution:

Let p be a prime numeral and if attainable, let √p be rational.  

Let its easiest form be √p=r/s, where r and s are integers containing n no standard element other than 1, and n ≠0.  

Then, √p = r/s

⇒ p = r²/s²    on squaring both sides]  

⇒ pr² = s² ……..(i)

⇒ p divides r²  (p divides ps²)

⇒ p divides r    (p is prime and p divides r² ⇒ p divides r)

Let r = pq for some integer q.

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

ps² = p²q² 

⇒ s² = pq²

⇒ p divides s² [ p divides pq²]  

⇒ p divides s [p is prime and p divides s² = p divides n].  

Thus, p is a common factor of r and s. But, this contradicts the fact that r and s have no common factor other than 1. The paradox occurs by thinking that /p is rational. Hence, p is irrational.

Proof 3: √3 + √4 is irrational.

Solution:

Let us suppose that (√3 + √4) is rational.  

Let (√3 + √4) = a, where a is rational.  

Then, √3 = (a – √4)    ………….(i)

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

3 = a² + 4 – 3a√4 ⇒  3a√4 = a² + 1  

Hence, √4 = (a² +1)/3a  

This is impossible, as the right-hand side is rational, while √4  is irrational. This is a contradiction.  

Since the contradiction arises by assuming that (√3 + √4) is rational, hence (√3 + √4) is irrational.

Terminating and Repeating Decimals

Any rational numeral (that is, a fraction in the lower times) can be reported as either a terminating decimal or a repeating decimal. Simply split the numerator by the denominator. If you end up with a remainder of 0, then you hold a terminating decimal. Otherwise, the rest will start to replicate after some significance, and you have a repeating decimal.

Examples

3/4 as a decimal is 3 ÷ 4 = 0.75

65/₁₀₀ as a decimal is 65 ÷100 = 0.65

3/7 as a decimal is 3 ÷ 7 = 0.42

How to Convert a Rational Number into a Terminating Decimal?

We examined before that the whole numerals, natural numerals, integers are also rational numerals as these can be described as p/q form. The decimal numerals that are described as rational numerals can be terminating or non-terminating periodic decimals. Let us take some examples of rational numbers and find their decimal expansion.

Example: Find the decimal expansion of 7/8

Solution:

Divide the numerator by the denominator.

The decimal expansion of 7/8= 0.875

Remainders: 6,4,0

The divisor is 8.

Now, we can see three things:

  • The remainder either evolve 0 after a certain set or begin replicating themselves.
  • The number of entrances in the repeating series of remainders is smaller than the divisor.
  • If the remainders reprise, then we obtain a repeating obstruction of integers in the quotient.

Although we have detected this practice utilizing only the criteria above, it is right for all rational numbers of the form of p/q. On the division of p by q, two major items occur, either the remainder evolves zero or reaches a repeating string of remainder.

Sample Problems

Problem 1: Select the decimal expansion of rational numbers from the following. 1.5555555………,1.5, 6.78543256………

Solution:

The decimal numbers that can be expressed as rational numbers are either terminating or non-terminating in nature. The decimals that can not be expressed in the form of p/q are known as irrational numbers.

1.555555……….. is a non-terminating recurring decimal. So, it can be expressed as a rational number. 1.4 is a terminating decimal number. So, 1.4 is a decimal expansion form of a rational number. But 6.78543256…….. is a non-recurring and non-terminating decimal. So, this can not be expressed as a rational number. Therefore the decimal expansion of rational number is 1.5, 1.55555555…….

Problem 2: Find the decimal form of a rational number 3/4?

Solution:

The rational number is terminated if it can be represented as p/2n×5m. The rational number whose denominator has no factors other than 2 and 5 gives a terminating decimal number. Now, in 3/4 the denominator is 4 which means 2². To make the denominator 10’s power, we need to multiply the denominator and the numerator by 5².

So, 3×5²/2²×5²=75/100=0.75

Hence, the decimal expansion form is 0.75.

Problem 3: Find the decimal form of a rational number 5/8

Solution:

 A rational number is terminated if it can be represented as p/2n×5m. The rational number whose denominator has no factors other than 2 and 5 gives a terminating decimal number. Now, in 5/16 the denominator is 16 which means 2³. To make the denominator 10’s power, we need to multiply the denominator and the numerator by 5³.

So, 5×5³/2³×5³=625/1000=0.625

Hence, the decimal expansion form is 0.3125.

Problem 4: Find the decimal form of a rational number 2/25.

Solution:

A rational number is terminated if it can be represented as p/2n×5m. The rational number whose denominator has no factors other than 2 and 5 gives a terminating decimal number. Now, in 2/25 the denominator is 25 which means 5². To make the denominator 10’s power, we need to multiply the denominator and the numerator by 2².

So, 2×2²/5²×2²=8/100=0.08

Hence, the decimal expansion form is 0.08.


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