有理数的十进制展开

在数字系统中，实数只是有理数和无理数的组合。通常，所有算术运算都可以在这些数字上执行，并且它们也可以在数字行中表示。因此，在本文中，我们讨论一些有理数和无理数及其证明。

有理数

p / q形式的数字(其中p和q是整数，q≠0)称为有理数。

例子：

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.

无理数

以十进制形式表示的数字可以表示为非终止和非重复的十进制数，称为无理数。

例子：

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.

有理数的十进制展开的性质

• 定理1：令x为最简单形式为p / q的有理数，其中p和q为整数，且q≠0。然后，仅当q的形式为(2 ^m x 5 ⁿ )时，x是终止小数。一些非负整数m和n。
• 定理2：令x为最简单形式为p / q的有理数，其中p和q为整数，且q≠0。然后，如果q≠(2 ^m x 5 ⁿ )，则x为无终止重复小数。
• 定理2：设x为有理数，其最简单形式为p / q，其中p和q为整数，q = 2 ^m x 5 ^n，则x具有终止的十进制扩展。

证明1：√2不合理

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 = p²/q²    (on squaring both sides)

⇒ 2q² = p²    ……..(i)

⇒ 2 divides p²   (2 divides 2q² ) 

⇒ 2 divides p   (2 is prime and divides p² ⇒ 2 divides p)

Let p = 2r for some integer r.

putting p = 2r in eqn (i)

2q² = 4r² 

⇒ q²= 2r²

⇒ 2 divides q²   (2 divides 2r² )

⇒ 2 divides p    (2 is prime and divides q² ⇒ 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.

证明2：素数的平方根是不合理的

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 = m²/n²         [on squaring both sides] 

⇒ pn² = m² ……..(i)

⇒ p divides m²  (p divides pn²)

⇒ p divides m    (p is prime and p divides m² ⇒ p divides m)

Let m = pq for some integer q.

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

pn² = p²q² 

⇒ n² = pq²

⇒ p divides n² [ p divides pq²] 

⇒ 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.

证明3：2 +√3是不合理的。

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.

证明4：√2+√3是不合理的。

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 = a² + 3 – 2a√3 ⇒  2a√3 = a² + 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.

标识终止小数

检查给定有理数是终止数还是重复小数让x是最简单形式为p / q的有理数，其中p和q是整数，q≠0。然后，

(i)仅当q为某些非负整数m和n^{的形式为(2 m} x 5 ^{n)时，x才是终止小数。}

^{(ii)如果q≠(2 m} x 5 ⁿ )，则x是一个无终止的重复小数。

例子

(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) = (2^m × 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) = (2^m × 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 ≠ (2^m × 5ⁿ) .

53 /343 is a non-terminating repeating decimal.