如何将有理数转换为终止小数？

在数字系统中，实数只是有理数和无理数的混合。众所周知，所有的算术功能都可以在这些数字上实现，它们也可以用数轴来表示。因此，在本文中，让我们研究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：

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.

无理数

当以十进制格式表示时，可以表示为非终止和非重复小数的数字被理解为无理数。

例子：

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.

有理数的小数展开式的字符

定理 1：设 x 是一个有理数，其最简单的形式是 p/q，其中 p 和 q 是整数并且 q ≠ 0。那么，只有当 q 的形式为 (2r x 5s) 时，x 才是终止小数-负整数 r 和 s。

定理 2：设 x 是一个有理数，其最简单的形式是 p/q，其中 p 和 q 是整数且 q ≠ 0。那么，如果 q ≠ (2r x 5s)，则 x 是一个非终止的重复小数。

定理 3：设 x 是一个有理数，其最简单的形式是 p/q，其中 p 和 q 是整数并且 q = 2r x 5s 那么 x 有一个小数展开式可以去除。

证明1：√3是无理数

解决方案：

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.

证明2：素数的平方根是无理数

解决方案：

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.

证明 3：√3 + √4 是非理性的。

解决方案：

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.

终止和重复小数

任何有理数（即较低时间的分数）都可以报告为终止小数或重复小数。只需将分子除以分母即可。如果你的余数为 0，那么你持有一个终止小数。否则，其余部分将在某个意义之后开始复制，并且您有一个重复的小数。

例子

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

如何将有理数转换为终止小数？

我们之前研究过整数、自然数、整数也是有理数，因为它们可以描述为 p/q 形式。被描述为有理数的十进制数字可以是终止或非终止周期小数。让我们举一些有理数的例子，并找出它们的十进制展开式。

示例：求 7/8 的小数展开式

解决方案：

Divide the numerator by the denominator.

The decimal expansion of 7/8= 0.875

Remainders: 6,4,0

The divisor is 8.

现在，我们可以看到三件事：

• 其余的要么在某个集合之后进化为 0，要么开始自我复制。
• 余数重复序列中的入口数小于除数。
• 如果余数重复出现，则我们在商中获得整数的重复阻塞。

尽管我们仅使用上述标准检测到这种做法，但它适用于 p/q 形式的所有有理数。在 p 除以 q 时，出现两个主要项目，余数演变为零或达到重复的余数字符串。

示例问题

问题1：从以下选择有理数的十进制展开式。 1.5555555…………,1.5, 6.78543256…………

解决方案：

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…….

问题 2：求有理数 3/4 的十进制形式？

解决方案：

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.

问题 3：求有理数 5/8 的十进制形式

解决方案：

 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.

问题 4：求有理数 2/25 的十进制形式。

解决方案：

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.