Course: 2.3 增加合理数字

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增加合理数字

Addition of Rational Numbers
::增加合理数字

In the last Concept, we learned how to represent numbers on a number line . To add numbers on a number line, we start at the position of the first number, and then move to the right by a number of units equal to the second number.
::在最后一个概念中, 我们学会了如何在数字行上代表数字。要在数字行上添加数字, 我们从第一个数字的位置开始, 然后由数个与第二个数字相等的单位向右移动。

Representing Sums on a Number Line
::数字行的合计数

1. Represent the sum $\begin{aligned} - 2 + 3 \end{aligned}$ on a number line.
::1. 在数字行中代表总和-2+3。

We start at the number -2, and then move 3 units to the right. We thus end at +1.
::我们从数字 - 2 开始,然后将3个单位移到右边。因此,我们以+1结束。

$\begin{aligned} - 2 + 3 = 1 \end{aligned}$

2. Represent the sum 2 - 3 on a number line.
::2. 数字行2-3之和。

Subtracting a number is basically just adding a negative number . Instead of moving to the right, we move to the left. Starting at the number 2, and then moving 3 to the left, means we end at -1.
::减法一个数字基本上是增加一个负数。我们不是向右移动,而是向左移动。从数字2开始,然后向左移动3,意味着我们以-1结束。

$\begin{aligned} 2 - 3 = - 1 \end{aligned}$

Adding and Subtracting Rational Numbers
::添加和减去理性数字

When we add or subtract two fractions, the denominators must match before we can find the sum or difference . We have already seen how to find a common denominator for two .
::当我们增加或减去两个分数时,分母必须匹配,才能找到总和或差数。我们已经看到如何为两个分母找到一个共同分母。

Simplify $\begin{aligned} \frac{3}{5} + \frac{1}{6} \end{aligned}$ .
::简化 35+16。

To combine these fractions, we need to rewrite them over a common denominator. We are looking for the lowest common denominator (LCD). We need to identify the lowest common multiple or least common multiple (LCM) of 5 and 6. That is the smallest number that both 5 and 6 divide into evenly (that is, without a remainder).
::要合并这些分数,我们需要用一个共同分母重写它们。我们正在寻找最小的公分母(LCD ) 。我们需要确定5和6的最小公分数(LCM ) 。这是5和6均分的最小数字(即没有剩余部分 ) 。

The lowest number that 5 and 6 both divide into evenly is 30. The LCM of 5 and 6 is 30, so the lowest common denominator for our fractions is also 30.
::5个和6个两者均分的最低数字是30个,5个和6个的LCM是30个,因此我们分数的最低共同标准也是30个。

We need to rewrite our fractions as new so that the denominator in each case is 30.
::我们需要重写我们分数的新版, 这样每个案例的分母是30。

If you think back to our idea of a cake cut into a number of slices, $\begin{aligned} \frac{3}{5} \end{aligned}$ means 3 slices of a cake that has been cut into 5 pieces. You can see that if we cut the same cake into 30 pieces (6 times as many) we would need 6 times as many slices to make up an equivalent fraction of the cake—in other words, 18 slices instead of 3.
::如果你回想一下我们把蛋糕切成几个片子的想法,35意味着一个被切成五个片子的蛋糕的3片。你可以看到,如果我们把同一个蛋糕切成30块(6倍于6倍),我们需要6倍的切成一个相当部分的蛋糕——换句话说,18片而不是3片。

$\begin{aligned} \frac{3}{5} \end{aligned}$ is equivalent to $\begin{aligned} \frac{18}{30} \end{aligned}$ .
::35等于1830年。

By a similar argument, we can rewrite the fraction $\begin{aligned} \frac{1}{6} \end{aligned}$ as a share of a cake that has been cut into 30 pieces. If we cut it into 5 times as many pieces, we need 5 times as many slices.
::类似的论点是,我们可以把第16部分改写成蛋糕的一部分,蛋糕被切成30块。如果我们把它切成5倍的一块,我们需要5倍的切片。

$\begin{aligned} \frac{1}{6} \end{aligned}$ is equivalent to $\begin{aligned} \frac{5}{30} \end{aligned}$ .
::16等于530。

Now that both fractions have the same denominator, we can add them. If we add 18 pieces of cake to 5 pieces, we get a total of 23 pieces. 23 pieces of a cake that has been cut into 30 pieces means that our answer is $\begin{aligned} \frac{23}{30} \end{aligned}$ .
::现在两个分数都有相同的分母, 我们可以添加它们。如果我们把18块蛋糕加到5块, 我们得到总共23块。 23块蛋糕被切成30块意味着我们的答案是2330。

\begin{aligned} \frac{3}{5} + \frac{1}{6} = \frac{18}{30} + \frac{5}{30} = \frac{23}{30} \end{aligned}

Notice that when we have fractions with a common denominator, we add the numerators but we leave the denominators alone . Here is this information in algebraic terms.
::请注意,当我们拥有带有共同分母的分分数时, 我们添加了分子, 但是我们只留下这些分母。这是用代数表示的信息。

When adding fractions: $\begin{aligned} \frac{a}{c} + \frac{b}{c} = \frac{a + b}{c} \end{aligned}$
::添加分数时: ac+bc=a+bc

So far, we’ve only dealt with examples where it’s easy to find the least common multiple of the denominators. With larger numbers, it isn’t so easy to be sure that we have the LCD . We need a more systematic method. In the next example, we will use the method of prime factors to find the least common denominator .
::到目前为止,我们只处理了很容易找到最小常见的分母数数的例子。数量越多,就越不容易确定我们是否拥有LCD。我们需要一种更系统的方法。在下一个例子中,我们将使用主要因素的方法来找到最小的共同分母。

Subtracting Rational Numbers
::正在减减理性数字

Simplify $\begin{aligned} \frac{29}{90} - \frac{13}{126} \end{aligned}$ .
::简化2990-13126。

To find the lowest common multiple of 90 and 126, we first find the prime factors of 90 and 126. We do this by continually dividing the number by factors until we can’t divide any further. You may have seen a factor tree before.
::为了找到90和126这一最低公数,我们首先发现90和126这两个主要因素,我们这样做的方式是不断将数字除以因数,直到我们无法再进一步分割。你可能见过一个因数树。

The factor tree for 90 looks like this:
::90 的因子树看起来是这样的:

The factor tree for 126 looks like this:
::126的系数树看起来是这样的:

The LCM for 90 and 126 is made from the smallest possible collection of primes that enables us to construct either of the two numbers. We take only enough instances of each prime to make the number with the greater number of instances of that prime in its factor tree.
::90和126的LCM是由最小的质数集合制成的,它使我们能够构建两个数字中的任何一个。我们只使用每个质数的足够实例来使质数在其因子树上出现较多的质数。

Prime	Factors in 90	Factors in 126	We Need
2	1	1	1
3	2	2	2
5	1	0	1
7	0	1	1

So we need one 2, two 3’s, one 5 and one 7. That gives us $\begin{aligned} 2 \cdot 3 \cdot 3 \cdot 5 \cdot 7 = 630 \end{aligned}$ as the lowest common multiple of 90 and 126. So 630 is the LCD for our calculation.
::因此我们需要一个 2 2 3 3 、 1 5 和 1 7 。这让我们得到 2 3 3 3 5 5 7= 630 的最低公分数 90 和 126 。所以 6 3 3 3 3 3 和 1 7 是 6 3 3 5 7 = 630 。

90 divides into 630 seven times (notice that 7 is the only factor in 630 that is missing from 90), so $\begin{aligned} \frac{29}{90} = \frac{7 \cdot 29}{7 \cdot 90} = \frac{203}{630} \end{aligned}$ .
::90 等于 630 7次(注意 7 是 630 中从 90 缺的唯一的因数), 2990 = 729790 = 203630 。

126 divides into 630 five times (notice that 5 is the only factor in 630 that is missing from 126), so $\begin{aligned} \frac{13}{126} = \frac{5 \cdot 13}{5 \cdot 126} = \frac{65}{630} \end{aligned}$ .
::126人分为630人5次(注意在126人失踪的630人中,仅有5人5个),因此13126=5135126=65630。

Now we complete the problem: $\begin{aligned} \frac{29}{90} - \frac{13}{126} = \frac{203}{630} - \frac{65}{630} = \frac{138}{630} \end{aligned}$ .
::现在我们解决了问题:2990-13126=203630-65630=138630。

This fraction simplifies . To be sure of finding the simplest form for $\begin{aligned} \frac{138}{630} \end{aligned}$ , we write out the prime factors of the numerator and denominator. We already know the prime factors of 630. The prime factors of 138 are 2, 3 and 23.
::此分数简单化。要确定要找到 138630 中最简单的形式, 我们写出分子和分母的质因数。我们已经知道 630 的质因数。 138 的质因数是 2、 3 和 23 。

$\begin{aligned} \frac{138}{630} = \frac{2 \cdot 3 \cdot 23}{2 \cdot 3 \cdot 3 \cdot 5 \cdot 7} \end{aligned}$ ; one factor of 2 and one factor of 3 cancels out, leaving $\begin{aligned} \frac{23}{3 \cdot 5 \cdot 7} \end{aligned}$ or $\begin{aligned} \frac{23}{105} \end{aligned}$ as our answer.
::138630 = 233232232323357; 2 和 33357的一个因数取消, 剩下的233_517或23105作为我们的答复。

Identify and Apply Properties of Addition
::添加属性的识别和应用

Three mathematical properties which involve addition are the commutative, associative, and the additive identity properties .
::需要添加的三种数学属性是通量、联运和添加特性特性。

Commutative property : When two numbers are added, the sum is the same even if the order of the items being added changes.
::动产:如果加上两个数字,即使所添加项目的顺序发生变化,总和也是一样的。

Example: $\begin{aligned} 3 + 2 = 2 + 3 \end{aligned}$
::示例:3+2=2+3

Associative Property : When three or more numbers are added, the sum is the same regardless of how they are grouped.
::共有财产:在增加三个或三个以上数字时,总和是相同的,而不论其分类方式如何。

Example: $\begin{aligned} (2 + 3) + 4 = 2 + (3 + 4) \end{aligned}$
::示例2+3)+4=2+(3+4)

Additive Identity Property : The sum of any number and zero is the original number.
::补充身份财产:任何数字和零的总和为原始数字。

Example: $\begin{aligned} 5 + 0 = 5 \end{aligned}$
::示例:5+0=5

Example
::示例示例示例示例

Example 1
::例1

Simplify $\begin{aligned} \frac{1}{3} - \frac{1}{9} \end{aligned}$ .
::简化13-19。

The lowest common multiple of 9 and 3 is 9, so 9 is our common denominator. That means we don’t have to alter the second fraction at all.
::9 和 3 的最低公数是 9 和 9 , 所以 9 是我们的共同分母。这意味着我们根本不需要改变第二分数。

3 divides into 9 three times, so $\begin{aligned} \frac{1}{3} = \frac{3 \cdot 1}{3 \cdot 3} = \frac{3}{9} \end{aligned}$ . Our sum becomes $\begin{aligned} \frac{3}{9} - \frac{1}{9} \end{aligned}$ . We can subtract fractions with a common denominator by subtracting their numerators, just like adding. In other words:
::3 分为 9 三次, 所以 13= 3133= 39。我们的金额是 39- 19 。我们可以用一个共同的分母来减减分数, 就像加上一样, 来减分数。换句话说 :

When subtracting fractions: $\begin{aligned} \frac{a}{c} - \frac{b}{c} = \frac{a - b}{c} \end{aligned}$
::减去分数时: ac-bc=a-bc

$\begin{aligned} \frac{1}{3} - \frac{1}{9} = \frac{2}{9} \end{aligned}$

Review
::回顾

Write the sum that the following moves on a number line represent.
::写入数字行的下列移动代表的和。

For 2-7, add the following rational numbers. Write each answer in its simplest form .
::2 - 7, 添加以下合理数字。以最简单的形式写下每个答案。

$\begin{aligned} \frac{3}{7} + \frac{2}{7} \end{aligned}$
$\begin{aligned} \frac{3}{10} + \frac{1}{5} \end{aligned}$
$\begin{aligned} \frac{5}{16} + \frac{5}{12} \end{aligned}$
$\begin{aligned} \frac{3}{8} + \frac{9}{16} \end{aligned}$
$\begin{aligned} \frac{8}{25} + \frac{7}{10} \end{aligned}$
$\begin{aligned} \frac{1}{6} + \frac{1}{4} \end{aligned}$

For 8-14, subtract the following rational numbers. Be sure that your answer is in the simplest form .
::8-14 时, 请减去以下的理性数字。请确定您的答案是最简单的。

$\begin{aligned} \frac{3}{4} - \frac{1}{3} \end{aligned}$
$\begin{aligned} \frac{15}{11} - \frac{9}{7} \end{aligned}$
$\begin{aligned} \frac{2}{13} - \frac{1}{11} \end{aligned}$
$\begin{aligned} \frac{7}{27} - \frac{9}{39} \end{aligned}$
$\begin{aligned} \frac{6}{11} - \frac{3}{22} \end{aligned}$
$\begin{aligned} \frac{13}{64} - \frac{7}{40} \end{aligned}$
$\begin{aligned} \frac{11}{70} - \frac{11}{30} \end{aligned}$

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

增加合理数字

Addition of Rational Numbers ::增加合理数字

Representing Sums on a Number Line ::数字行的合计数

Adding and Subtracting Rational Numbers ::添加和减去理性数字

Subtracting Rational Numbers ::正在减减理性数字

Identify and Apply Properties of Addition ::添加属性的识别和应用

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)