Learning Objectives

- Subtract two or more real numbers.
- Simplify combinations that require both addition and subtraction of real numbers.
- Solve application problems requiring subtraction of real numbers.

Subtraction and addition are closely related. They are called **inverse operations**, because one "undoes" the other. So, just as with integers, you can rewrite subtraction as addition to subtract real numbers.

Inverse operations, such as addition and subtraction, are a key idea in algebra. Suppose you have $10 and you loan a friend $5. An hour later, she pays you back the $5 she borrowed. You are back to having $10. You could represent the transaction like this:

10-5+5=10.

This works because a number minus itself is 0.

( 3-3=0 quad 63.5-63.5=0 quad 39,283-39,283=0)

So, adding a number and then subtracting the same number is like adding 0.

Thinking about this idea in terms of **opposite** numbers, you can also say that a number plus its opposite is also 0. Notice that each example below consists of a positive and a negative number pair added together.

( 3+(-3)=0 quad-63.5+63.5=0 quad 39,283+(-39,283)=0)

Two numbers are **additive inverses** if their sum is 0. Since this means the numbers are opposites (same absolute value but different signs), "additive inverse" is another, more formal term for the opposite of a number. (Note that 0 is its own additive inverse.)

You can use the additive inverses or opposites to rewrite subtraction as addition. If you are adding two numbers with different signs, you find the difference between their absolute values and keep the sign of the number with the greater absolute value.

When the greater number is positive, it's easy to see the connection.

( 13+(-7)=13-7)

Both equal 6.

Let’s see how this works. When you add positive numbers, you are moving forward, facing in a positive direction.

When you subtract positive numbers, you can imagine moving backward, but still facing in a positive direction.

Now let's see what this means when one or more of the numbers is negative.

Recall that when you add a negative number, you move forward, but face in a negative direction (to the left).

How do you subtract a negative number? First face and move forward in a negative direction to the first number, -2. Then continue facing in a negative direction (to the left), but move *backward *to subtract -3.

But isn’t this the same result as if you had added positive 3 to -2? -2+3=1.

## Supplemental Interactive Activity

Use the interactive number line below to find the answers to the following pairs of sums and differences, and compare the answers. You will have to specify both numbers and whether you are adding or subtracting.

( 3-4 ext { and } 3+(-4))

( 2-(-3) ext { and } 2+3)

( -1-5 ext { and }-1+(-5))

( -2-(-1) ext { and }-2+1)

In each addition problem, you move in one direction some distance forward. In the paired subtraction problem, you move in the *opposite* direction the same distance backward. Notice how each gives you the same result!

To subtract a real number, you can rewrite the problem as adding the opposite (additive inverse).

Note, that while this always works, whole number subtraction is still the same. You can subtract 38-23 just as you have always done. Or, you could also rewrite it as

38+(-23). Both ways you will get the same answer.

38-23=38+(-23)=15.

It’s your choice in these cases.

Example

**Find 23-73.**

**Solution**

You can't use your usual method of subtraction, because 73 is greater than 23. | |

( 23+(-73)) | Rewrite the subtraction as adding the opposite. |

( egin{array}{c} |23|=23 ext { and }|-73|=73 73-23=50 end{array}) | The addends have different signs, so find the difference of their absolute values. |

( 23-73=-50) | Since ( |-73|>|23|), the final answer is negative. |

Example

**Find ( 382-(-93)).**

**Solution**

( 382+93) ( 382+93=475) | Rewrite the subtraction as adding the opposite. The opposite of -93 is 93. So, this becomes a simple addition problem. |

( 382-(-93)=475)

Another way to think about subtracting is to think about the distance between the two numbers on the number line. In the example above, 382 is to the *right* of 0 by 382 units, and -93 is to the *left* of 0 by 93 units. The distance between them is the sum of their distances to 0: 382+93.

Example

**Find ( 22 frac{1}{3}-x), when ( x=-frac{3}{5}).**

**Solution**

( 22 frac{1}{3}-left(-frac{3}{5} ight)) | Substitute ( -frac{3}{5}) for ( x) in the expression. |

( 22 frac{1}{3}+frac{3}{5}) | Rewrite the subtraction as adding the opposite. The opposite of ( -frac{3}{5}) is ( frac{3}{5}). |

( 22 frac{1 cdot 5}{3 cdot 5}+frac{3 cdot 3}{5 cdot 3}=22 frac{5}{15}+frac{9}{15}) ( 22 frac{5}{15}+frac{9}{15}=22 frac{14}{15}) | This is now just adding two rational numbers. Remember to find a common denominator when adding fractions. 3 and 5 have a common multiple of 15; change denominators of both fractions to 15 (and make the necessary changes in the numerator!) before adding. |

( 22 frac{14}{15})

Exercise

Find ( -32.3-(-16.3)).

- -48.6
- -16
- 16
- 48.6

**Answer**-48.6

Incorrect. You added -32.3 and -16.3. To subtract, change the problem to adding the opposite of -16.3, which gives -32.3+16.3. Then use the rules for adding two numbers with different signs. Since the difference between 32.3 and 16.3 is 16, and |-32.3|>|16.3|, the correct answer is -16.

-16

Correct. To subtract, change the problem to adding the opposite of -16.3, which gives -32.3+16.3. Then use the rules for adding two numbers with different signs. Since the difference between 32.3 and 16.3 is 16, and |-32.3|>|16.3|, the correct answer is -16.

16

Incorrect. You used the wrong sign. To subtract, change the problem to adding the opposite of -16.3, which gives -32.3+16.3. Then use the rules for adding two numbers with different signs. Since the difference between 32.3 and 16.3 is 16, and |-32.3|>|16.3|, the correct answer is -16.

48.6

Incorrect. You added the opposites of both numbers. To subtract, change the problem to adding the opposite of -16.3, which gives -32.3+16.3. Then use the rules for adding two numbers with different signs. Since the difference between 32.3 and 16.3 is 16, and |-32.3|>|16.3|, the correct answer is -16.

When you have more than two **real numbers** to add or subtract, work from left to right as you would when adding more than two whole numbers. Be sure to change subtraction to addition of the opposite when needed.

Example

**Find -23+16-(-32)-4+6.**

**Solution**

( egin{array}{r} {f-23+16}-(-32)-4+6 {f-7}-(-32)-4+6 end{array}) | Start with -23+16. The addends have different signs, so find the difference and use the sign of the addend with the greater absolute value. -23+16=-7. |

( egin{array}{r} {f -7-(-32)}-4+6 {f-7+32}-4+6 end{array}) | Now you have -7-(-32). Rewrite this subtraction as addition of the opposite. The opposite of -32 is 32, so this becomes -7+32, which equals 25. |

( {f25-4}+6) | You now have 25-4. You could rewrite this as an addition problem, but you don't need to. |

( f{21}+6) | Complete the final addition of 21+6. |

-23+16-(-32)-4+6=27

Exercise

Find 32-(-14)-2+(-82).

- -66
- -38
- 98
- 126

**Answer**-66

Incorrect. You probably subtracted -14 incorrectly. To subtract 32-(-14), write the subtraction as addition of the opposite, giving 32+14=46. Then subtract 2 to get 44, and add -82 to get the correct answer of -38.

-38

Correct. To subtract 32-(-14), write the subtraction as addition of the opposite, giving 32+14=46. Then subtract 2 to get 44, and add -82 to get -38.

98

Incorrect. You may have missed the negative signs on -14 and -82. To subtract 32-(-14), write the subtraction as addition of the opposite, giving 32+14=46. Then subtract 2 to get 44, and add -82 to get the correct answer of -38.

126

Incorrect. You probably subtracted -14 correctly, but you added 82 instead of -82 as the last step. To subtract 32-(-14), write the subtraction as addition of the opposite, giving 32+14=46. Then subtract 2 to get 44, and add -82 to get the correct answer of -38.

Situations that use negative numbers can require subtraction as well as addition. As you saw above, sometimes subtracting two positive numbers can give a negative result. You should be sure that a negative number makes sense in the problem.

Example

**Boston is, on average, 7 degrees warmer than Bangor, Maine. The low temperature on one cold winter day in Boston was 3 ^{o}F. About what low temperature would you expect Bangor to have on that day?**

**Solution**

If the temperature in Boston is ( x), the temperature in Bangor is ( x-7). | The phrase "7 degrees warmer" means you can subtract 7 degrees from Boston's temperature to estimate Bangor's temperature. (Note that you can also add 7 degrees to Bangor's temperature to estimate Boston's temperature. Be careful about which should have the greater number!) |

( x=3) | On that day, Boston's low was 3^{o}. |

Bangor's temperature is ( 3-7) | Substitute 3 for ( x) to get Bangor's temperature. |

( 3-7=3+(-7)) | Since 3<7, rewrite the subtraction problem as addition of the opposite. Add the numbers. Since one is positive and the other is negative, you find the difference of |-7| and |3|, which is 4. Since |-7|>|3|, the final sum is negative. |

You would expect the low temperature in Bangor, Maine to be -4^{o}F.

Example

**Everett paid several bills without balancing his checkbook first! When the last check he wrote was still to be deducted from his balance, Everett's account was already overdrawn. The balance was -$201.35. The final check was for $72.66, and another $25 will be subtracted as an overdraft charge. What will Everett's account balance be after that last check and the overdraft charge are deducted?**

**Solution**

( -201.35-72.66-25) | The new balance will be the existing balance of -$201.35, minus the check's amount and the overdraft charge. |

( egin{array}{r} -201.35-72.66-25 \ -201.35+(-72.66)-25 end{array}) | Start with the first subtraction, ( -201.35-72.66). Rewrite it as the addition of the opposite of 72.66. |

( -274.01-25) | Since the addends have the same signs, the sum is the sum of their absolute values (201.35+72.66) with the same sign (negative). |

( -274.01+(-25)) | Again, rewrite the subtraction as the addition of the opposite. |

( -274.01+(-25)=-299.01) | Add, by adding the sum of their absolute values and use the same sign as both addends. |

Everett’s account balance will be -$299.01.

Example

**One winter, Phil flew from Syracuse, NY to Orlando, FL. The temperature in Syracuse was -20 ^{o}F. The temperature in Orlando was 75^{o}F. What was the difference in temperatures between Syracuse and Orlando? **

**Solution**

( 75-(-20)) | To find the difference between the temperatures, you need to subtract. We subtract the ending temperature from the beginning temperature to get the change in temperature. |

( 75+20) | Rewrite the subtraction as adding the opposite. The opposite of -20 is 20. |

( 75+20=95) | There is a 95 degree difference between 75^{o} and -20^{o}. |

The difference in temperatures is 95 degrees.

Exercise

Louise noticed that her bank balance was -$33.72 before her paycheck was deposited. After the check had been deposited, the balance was $822.98. No other deductions or deposits were made. How much money was she paid?

**Answer**$856.70. The amount she was paid is the difference between the two balances: ( 822.98-(-33.72)). This is the same as ( 822.98+33.72), or 856.70.

Subtracting a number is the same as adding its opposite (also called its additive inverse). To subtract, you can rewrite the subtraction as adding the opposite and then use the rules for the addition of real numbers.

## How to Subtract Two or More Numbers in Excel

### What to Know

- The basic subtraction formula is
**=(cell location) - (cell location)**. - The subtraction sign is denoted by the dash ( - ).
- More complicated problems need a good understanding of how Excel handles order of operations.

This article covers how to handle simple and complicated subtraction formulas in Excel.

The instructions in this article apply to Excel 2019, Excel 2016, Excel 2013, Excel 2010, Excel for Mac, and Excel Online.

## How to Subtract a Mixed Number from Another?

- Divide the numerator by the denominator
- The whole part of the quotient is the whole number of the mixed number
- The reminder is the new numerator of the proper fraction
- The denominator of the proper fraction is equal to the denominator of the improper fraction.

- Multiply the denominator of the proper fraction by the whole number in the mixed number and add it to its numerator
- The denominator of the improper fraction is equal to the denominator of the proper fraction of the mixed number.

**When denominators of proper fractions of mixed numbers are equal**

**When denominators of proper fractions of mixed numbers are different**

- Convert mixed numbers to corresponding improper fractions
- Find the LCM of denominators of derived improper fractions
- Rewrite these fractions over the LCM
- Subtract the second numerator from the firs one
- The result is the difference of numerators over the LCM
- Simplify the result if needed.

### Real World Problems Using Mixed Numbers Subtraction

Mixed numbers are useful in counting whole things and parts of these things together. It is used primarily in measurement. Especially of interest are mixed numbers whose denominator of the fractional part is a power of two. They are commonly used with U.S. customary units such as inches, pounds, etc. For instance, $1 <
m inch>=2frac<54><100>
m

### Mixed Numbers Subtraction Practice Problems

**Practice Problem 1 :**

There are $36frac 14$ boxes of tomatoes in a truck. Farmer sold $21frac 37$ boxes of tomatoes. How many boxes of tomatoes left in the truck?

**Practice Problem 2 :**

Mitchell's ice cream recipe calls for $3 frac 35$ caps of sugar and Ann's recipe calls for $1 frac38$ caps of sugar. How many more caps of sugar are used in Mitchell's recipe than in Ann's recipe?

The mixed number subtraction calculator, formula, example calculation (work with steps), real world problems and practice problems would be very useful for grade school students (K-12 education) to understand the subtraction of two or more numbers represented as mixed numbers. Using this concept they can be able to solve complex algebraic problems and equations.

## Subtracting Integers by Adding the Opposite

Videos, solutions, worksheets, and songs to help Grade 6 students learn how to subtract integers by "adding the opposite".

The following figures show how to subtract integers by adding the opposite. Scroll down the page for more examples and solutions.

To find the opposite of an integer, we just change its sign. If it is negative, change to positive. If it is positive change to negative.

To subtract an integer, we can rewrite it as adding the opposite.

For eg.

3 &minus 2 = 3 + (&minus2)

3 &minus (&minus2) = 3 + (+2)

Then, we use the same rules as adding integers.

The rules for adding integers are:

1. If the signs are the same, add the absolute value of the numbers and keep the sign.

2. If the signs are different,

&bull subtract the smaller absolute value from the greater absolute value.

&bull the sign will follow the number with the greater absolute value.

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

## Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1

Question 1.

Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.

Solution:

i. Denominator = 5 = 1 x 5

Since, 5 is the only prime factor denominator.

the decimal form of the rational number (frac < 13 >< 5 >) will be terminating type.

ii. Denominator = 11 = 1 x 11

Since, the denominator is other than prime factors 2 or 5.

∴ the decimal form of the rational number (frac < 2 >< 11 >) will be non-terminating recurring type.

iii. Denominator = 16

= 2 x 2 x 2 x 2

Since, 2 is the only prime factor in the denominator.

∴ the decimal form of the rational number (frac < 29 >< 16 >) will be terminating type.

iv. Denominator = 125

= 5 x 5 x 5

Since, 5 is the only prime factor in the denominator.

the decimal form of the rational number (frac < 17 >< 125 >) will be terminating type.

v. Denominator = 6

= 2 x 3

Since, the denominator is other than prime factors 2 or 5.

∴ the decimal form of the rational number (frac < 11 >< 6 >) will be non-terminating recurring type.

Question 2.

Write the following rational numbers in decimal form.

Solution:

i. (frac < -5 >< 7 >)

ii. (frac < 9 >< 11 >)

iii. √5

iv. (frac < 121 >< 13 >)

v. (frac < 29 >< 8 >)

Question 3.

Write the following rational numbers in (frac < p >< q >) form.

Solution:

i. Let x = (0.dot < 6 >) …(i)

∴ x = 0.666…

Since, one number i.e. 6 is repeating after the decimal point.

Thus, multiplying both sides by 10,

10x = 6.666…

∴ 10 x 6.6 …(ii)

Subtracting (i) from (ii),

10x – x = 6.6 – 0.6

∴ 9x = 6

ii. Let x = (0.overline < 37 >)

∴ x = 0.3737…

Since, two numbers i.e. 3 and 7 are repeating after the decimal point.

Thus, multiplying both sides by 100,

100x = 37.3737……

∴ 100x = (37.overline < 37 >) ……(ii)

Subtracting (i) from (ii),

100x – x = (37.overline < 37 >) – (0.overline < 37 >)

∴ 99x = 37

iii. Letx = (3.overline < 17 >) …(i)

∴ x = 3.1717…

Since, two numbers i.e. 1 and 7 are repeating after the decimal point.

Thus, multiplying both sides by 100,

100x = 317.1717…

∴ 100x= 317.17 …(ii)

Subtracting (i) from (ii),

100x – x = (317.overline < 17 >) – (3.overline < 17 >)

∴ 99x = 314

iv. Let x = (15.overline < 89 >) …….. (i)

∴ x = 15.8989…

Since, two numbers i.e. 8 and 9 are repeating after the decimal point.

Thus, multiplying both sides by 100,

100x= 1589.8989…

∴ 100x = (1589.overline < 89 >) …(ii)

Subtracting (i) from (ii),

100x – x = (1589.overline < 89 >) – (15.overline < 89 >)

∴ 99x = 1574

v. Let x = (2.overline < 514 >)

∴ x = 2.514514…

Since, three numbers i.e. 5, 1 and 4 are repeating after the decimal point.

Thus, multiplying both sides by 1000,

1000x = 2514.514514…

1000x = (2514.overline < 514 >) ….(ii)

Subtracting (i) from (ii),

1000x – x = (2514.overline < 514 >) – (2.overline < 514 >)

∴ 999x = 2512

Question 1.

How to convert 2.43 in (frac < p >< q >) form ? (Textbook pg. no. 20)

Solution:

Let x = 2.43

In 2.43, the number 4 on the right side of the decimal point is not recurring.

So, in order to get only recurring digits on the right side of the decimal point, we will multiply 2.43 by 10.

∴ 10x = 24.3 …(i)

∴ 10x = 24.333…

Here, digit 3 is the only recurring digit. Thus, by multiplying both sides by 10, 100x = 243.333…

∴ 100x= 243.3 …(ii)

Subtracting (i) from (ii),

100x – 10x = 243.3 – 24.3

∴ 90x = 219

## Mathematics_part_ _i_(solutions) for Class 9 Math Chapter 2 - Real Numbers

Mathematics_part_ _i_(solutions) Solutions for Class 9 Math Chapter 2 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 9 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics_part_ _i_(solutions) Book of Class 9 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics_part_ _i_(solutions) Solutions. All Mathematics_part_ _i_(solutions) Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Page No 21:

#### Question 1:

Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.

i 13 5 ii 2 11 iii 29 16 iv 17 125 v 11 6

#### Answer:

⇒ The denominator is in the form of 2 m × 5 n , where *m* and *n* are non-negative integers.

So, the decimal form of 13 5 will be terminating type.

⇒ The denominator is not in the form of 2 m × 5 n , where *m* and *n* are non-negative integers.

So, the decimal form of 2 11 will be non-terminating recurring type.

⇒ The denominator is in the form of 2 m × 5 n , where *m* and *n* are non-negative integers.

So, the decimal form of 29 16 will be terminating type.

⇒ The denominator is in the form of 2 m × 5 n , where *m* and *n* are non-negative integers.

So, the decimal form of 17 125 will be terminating type.

⇒ The denominator is not in the form of 2 m × 5 n , where *m* and *n* are non-negative integers.

So, the decimal form of 11 6 will be non-terminating recurring type.

#### Page No 21:

#### Question 2:

Write the following rational numbers in decimal form.

i 127 200 ii 25 99 iii 23 7 iv 4 5 v 17 8

#### Answer:

i 127 200 = 127 200 × 5 5 = 635 1000 = 0 . 635

ii 25 99 = 4 4 × 25 99 = 1 4 × 100 99 = 1 4 × 1 . 010101 . . . = 0 . 2525 . . . = 0 . 25 ¯

iii 23 7 = 3 . 2857142857 . . . = 3 . 285714 ¯

v 17 8 = 17 8 × 125 125 = 2125 1000 = 2 . 125

#### Page No 21:

#### Question 3:

Write the following rational numbers in p q form

#### Answer:

i Let x = 0 . 6 ° . . . 1 x = 0 . 666 . . . Multiplying both sides by 10 , we get 10 x = 6 . 666 . . . . . . 2 Subtracting 1 from 2 , we get 9 x = 6 ∴ x = 6 9 So , 0 . 6 ° = 2 3

ii Let x = 0 . 37 ¯ . . . 1 Multiplying both sides by 100 , we get 100 x = 37 . 37 ¯ . . . 2 Subtracting 1 from 2 , we get 99 x = 37 ∴ x = 37 99 So , 0 . 37 ¯ = 37 99

iii Let x = 3 . 17 ¯ . . . 1 Multiplying both sides by 100 , we get 100 x = 317 . 17 ¯ . . . 2 Subtracting 1 from 2 , we get 99 x = 314 ∴ x = 314 99 So , 3 . 17 ¯ = 314 99

iv Let x = 15 . 89 ¯ . . . 1 Multiplying both sides by 100 , we get 100 x = 1589 . 89 ¯ . . . 2 Subtracting 1 from 2 , we get 99 x = 1574 ∴ x = 1574 99 So , 3 . 17 ¯ = 1574 99

v Let x = 2 . 514 ¯ . . . 1 Multiplying both sides by 1000 , we get 1000 x = 2514 . 514 ¯ . . . 2 Subtracting 1 from 2 , we get 999 x = 2512 ∴ x = 2512 999 So , 2 . 514 ¯ = 2512 999

#### Page No 25:

#### Question 1:

Show that 4 2 is an irrational number.

#### Answer:

Let us assume that 4 2 is a rational number.

⇒ 4 2 = p q , where *p* and *q* are the integers and *q* ≠ 0.

Since, *p*, *q* and 4 are integers. So, p 4 q is a rational number.

⇒ 2 is also a rational number.

but this contradicts the fact that 2 is an irrational number.

This contradiction has arisen due to the wrong assumption that 4 2 is a rational number.

Hence, 4 2 is an irrational number.

#### Page No 25:

#### Question 2:

Prove that 3 + 5 is an irrational number.

#### Answer:

Let us assume that 3 + 5 is a rational number.

⇒ 3 + 5 = p q , where *p* and *q* are the integers and *q* ≠ 0.

Since, *p*, *q* and 3 are integers. So, p - 3 q q is a rational number.

⇒ 5 is also a rational number.

but this contradicts the fact that 5 is an irrational number.

This contradiction has arisen due to the wrong assumption that 3 + 5 is a rational number.

Hence, 3 + 5 is an irrational number.

#### Page No 25:

#### Question 3:

Represent the numbers 5 and 10 on a number line .

#### Answer:

(i) Steps of construction for 5 :

Step 1: Draw a number line. Mark O as the zero on the number line.

Step 2: At point A, draw AB ⊥ OA such that AB = 1 unit.

Step 3: With point O as the centre and radius OB, draw an arc intersecting the number line at point P.

Thus, P is the point for 5 on the number line.

(ii) Steps of construction for 10 :

Step 1: Draw a number line. Mark O as the zero on the number line.

Step 2: At point A, draw AB ⊥ OA such that AB = 1 unit.

Step 3: With point O as the centre and radius OB, draw an arc intersecting the number line at point C.

## Setting up the Problems

The big symbol you will learn in subtraction is the **minus** sign. It's a little dash (-) that goes between the two numbers in the problem. The order of the values in a subtraction problem is very, very important. In addition, you could move the numbers around and get the same answer. If you move numbers in a subtraction problem, the answer will be wrong. You cannot reorder numbers when you subtract.

**Examples:**

3 + 9 = 12

9 + 3 = 12 (You can reorder in addition and get the same sum.)

9 - 3 = 6

3 - 9 = -6 (See how the answer is a negative number? If you put the numbers in the wrong order on a test you will get the wrong answer.)

10 - 5 - 3 = 2

3 - 10 - 5 = -12

5 - 3 - 10 = -8 (Don't rearrange the numbers in your subtraction problems!)

The numbers in a subtraction problem also have special names. You don't need to memorize them, just know that they have special names. The first value is the **minuend**. The second value (the one you are subtracting) is called the **subtrahend**. The answer in a subtraction problem is called the **difference**. Actually, you probably should remember that the answer to a subtraction problem is called the difference.

**Example:**

Problem: 9 - 3 = 6

Minuend: 9

Subtrahend: 3

Difference: 6

## Grouping Symbols and Exponents

In a computation where more than one operation is involved, grouping symbols help tell us which operations to perform first. The grouping symbols Parentheses, brackets, braces, and the fraction bar are the common symbols used to group expressions and mathematical operations within a computation. commonly used in algebra are:

( ) P a r e n t h e s e s [ ] B r a c k e t s < >B r a c e s F r a c t i o n b a r

All of the above grouping symbols, as well as absolute value, have the same order of precedence. Perform operations inside the innermost grouping symbol or absolute value first.

### Example 8

Perform the operations within the parentheses first.

2 − ( 4 5 − 2 15 ) = 2 − ( 4 5 ⋅ 3 3 − 2 15 ) = 2 − ( 12 15 − 2 15 ) = 2 − ( 10 15 ) = 2 1 ⋅ 3 3 − 2 3 = 6 − 2 3 = 4 3

### Example 9

Simplify: 5 − | 4 − ( − 3 ) | | − 3 | − ( 5 − 7 ) .

The fraction bar groups the numerator and denominator. Hence, they should be simplified separately.

5 − | 4 − ( − 3 ) | | − 3 | − ( 5 − 7 ) = 5 − | 4 + 3 | | − 3 | − ( − 2 ) = 5 − | 7 | | − 3 | + 2 = 5 − 7 3 + 2 = − 2 5 = − 2 5

If a number is repeated as a factor numerous times, then we can write the product in a more compact form using exponential notation The compact notation a n used when a factor *a* is repeated *n* times. . For example,

The base The factor *a* in the exponential notation a n . is the factor and the positive integer exponent The positive integer *n* in the exponential notation a n that indicates the number of times the base is used as a factor. indicates the number of times the base is repeated as a factor. In the above example, the base is 5 and the exponent is 4. Exponents are sometimes indicated with the caret (^) symbol found on the keyboard, 5^4 = 5*5*5*5. In general, if *a* is the base that is repeated as a factor *n* times, then

When the exponent is 2 we call the result a square The result when the exponent of any real number is 2. , and when the exponent is 3 we call the result a cube The result when the exponent of any real number is 3. . For example,

5 2 = 5 ⋅ 5 = 25 “ 5 s q u a r e d ” 5 3 = 5 ⋅ 5 ⋅ 5 = 125 “ 5 c u b e d ”

If the exponent is greater than 3, then the notation a n is read, “*a raised to the nth power*.” The base can be any real number,

( 2.5 ) 2 = ( 2.5 ) ( 2.5 ) = 6.25 ( − 2 3 ) 3 = ( − 2 3 ) ( − 2 3 ) ( − 2 3 ) = − 8 27 ( − 2 ) 4 = ( − 2 ) ( − 2 ) ( − 2 ) ( − 2 ) = 16 − 2 4 = − 1 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = − 16

Notice that the result of a negative base with an even exponent is positive. The result of a negative base with an odd exponent is negative. These facts are often confused when negative numbers are involved. Study the following four examples carefully:

( − 3 ) 4 = ( − 3 ) ( − 3 ) ( − 3 ) ( − 3 ) = + 81 ( − 3 ) 3 = ( − 3 ) ( − 3 ) ( − 3 ) = − 27

− 3 4 = − 1 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3 = − 81 − 3 3 = − 1 ⋅ 3 ⋅ 3 ⋅ 3 = − 27

The parentheses indicate that the negative number is to be used as the base.

### Example 10

Here − 1 3 is the base for both problems.

Use the base as a factor three times.

( − 1 3 ) 3 = ( − 1 3 ) ( − 1 3 ) ( − 1 3 ) = − 1 27

Use the base as a factor four times.

( − 1 3 ) 4 = ( − 1 3 ) ( − 1 3 ) ( − 1 3 ) ( − 1 3 ) = + 1 81

**Try this!** Simplify:

## MathHelp.com

Let's return to the first example from the previous page: " 9 &ndash 5 " can also be written as " 9 + (&ndash5) ". Graphically, it would be drawn as "an arrow from zero to nine, and then a 'negative' arrow five units long":

Now look back at that subtraction you couldn't do: 5 &ndash 9 . Because you now have negative numbers off to the left of zero, you also now have the "space" to complete this subtraction. View the subtraction as adding a negative 9 that is, draw an arrow from zero to five, and then a "negative" arrow nine units long:

. or, which is the same thing:

Of course, this method of counting off your answer on a number line won't work so well if you're dealing with larger numbers. For instance, think about doing " 465 &ndash 739 ". You certainly don't want to use a number line for this. However, since 739 is larger than 465 , you know that the answer to " 465 &ndash 739 " has to be negative, because "minus 739 " will take you somewhere to the left of zero. But how do you figure out *which* negative number is the answer?

Look again at " 5 &ndash 9 ". You know now that the answer will be negative, because you're subtracting a bigger number than you'd started with (the nine is bigger than the five). The easiest way of dealing with this is to do the subtraction "normally" (with the smaller number being subtracted from the larger number), and then put a "minus" sign on the answer: 9 &ndash 5 = 4 , so 5 &ndash 9 = &ndash4 . This works the same way for bigger numbers (and is much simpler than trying to draw the picture): since 739 &ndash 465 = 274 , then 465 &ndash 739 = &ndash274 .

Adding two negative numbers is easy: you're just adding two "negative" arrows, so it's just like "regular" addition, but in the opposite direction. For instance, 4 + 6 = 10 , and &ndash4 &ndash 6 = &ndash4 + (&ndash6) = &ndash10 . But what about when you have lots of both positive and negative numbers?

#### Simplify 18 &ndash (&ndash16) &ndash 3 &ndash (&ndash5) + 2

Probably the simplest thing to do is convert everything to addition, group the positives together and the negatives together, combine, and simplify. It looks like this:

"Whoa! Wait a minute!" I hear you say. "How do you go from ' &ndash (&ndash16) ' to ' +16 ' in your first step? How did the 'minus of minus 16 ' turn into a 'plus 16 '?"

This is actually a fairly important concept, and, if you're asking, I'm assuming that your teacher's explanation didn't make much sense to you. So I won't give you a "proper" mathematical explanation of this "the minus of a minus is a plus" rule. Instead, here's a mental picture that I ran across years ago in an algebra newsgroup:

Imagine that you're cooking some kind of stew in a big pot, but you're not cooking on a stove. Instead, you control the temperature of the stew with magic cubes. These cubes come in two types: hot cubes and cold cubes.

If you add a hot cube (add a positive number) to the pot, the temperature of the stew goes up. If you add a cold cube (add a negative number), the temperature goes down. If you remove a hot cube (subtract a positive number), the temperature goes down. And if you remove a cold cube (subtract a negative number), *the temperature goes UP!* That is, subtracting a negative is the same as adding a positive.

Now suppose you have some double cubes and some triple cubes. If you add three double-hot cubes (add three-times-positive-two), the temperature goes up by six. And if you remove two triple-cold cubes (subtract two-times-negative-three), you get the same result. That is, &ndash2(&ndash3) = + 6 .

Here's another analogy that I've seen. Letting "good" be "positive" and "bad" be "negative", you could say:

good things happening to good people: a good thing

good things happening to bad people: a bad thing

bad things happening to good people: a bad thing

bad things happening to bad people: a good thing

To give a specific example:

the family of four in the minivan gets home, safe and sound: a good thing

the drunk driver in the stolen car veering all over the road doesn't get caught and stopped: a bad thing

the family of four is killed by the drunk driver, while the drunk flees the scene without a scratch: a bad thing

the drunk driver is caught and locked up before he hurts anybody: a good thing

The analogies above aren't technical explanations or proofs, but I hope they make the "minus of a minus is a plus" and "minus times minus is plus" rules seem a bit more reasonable.

For whatever reason, it seems helpful to use the terms "plus" and "minus" instead, of "add, "subtract", "positive", and "negative". So, for instance, instead of saying "subtracting a negative", you'd say "minus-ing a minus". I have no idea why this is so helpful, but I do know that this verbal technique helped negatives "click" with me, too.

## Video-Lesson Transcript

In this lesson, we’re going to discuss how to add and subtract signed numbers. This is adding and subtracting positive and negative numbers.

Now, let’s get to add and subtract signed numbers.

Let’s start with adding like sign numbers. This might be the easiest.

Let’s start with a classic example.

This is the addition of two positive numbers.

We’ll start with and adding three more.

In the number line, we end up with .

Let’s see if we add negative numbers together.

So, let’s start with and add more.

In the number line, we end up with .

We end up with in the number line.

Now, let’s make sense out of this.

If we add two like sign numbers, we just have to add the numbers together and keep the sign, whether positive or negative.

Since the two numbers have the same sign, we just add them together and keep the negative sign.

Now, let’s go over adding unlike signs.

Let’s draw another number line.

Now, let’s have then add . So let’s move to the left.

Let’s start with then add . Let’s move to the right.

In the number line, we end up in .

So let’s start at in the number line and add . We get .

Let’s see what the pattern is.

When the signs are different, we find the difference between the numbers.

So the difference between and is . And it’s negative because the bigger number is negative.

In , the difference of the two numbers is . And since the bigger number is negative, the answer is negative.

Here in , since they have unlike signs we have to find the difference and the answer is positive because the bigger number is positive. So the answer is .

Let’s do another example without the number line.

Here, we find the difference between and which is . The bigger number is positive so our answer is also positive.

In adding unlike signs, we find the difference and we keep the sign of the larger number.

Let’s go over subtracting signed numbers.

We have , in the number line we end up with .

, we start at and subtract going to the left. Here, we end up with .

What if we have , start off at and we’re taking away. In the number line, we get .

Let’s say you have and we took away .

That’s a good thing. Maybe you owed someone and you took it away. You don’t owe someone anymore. The person said, “forget about it”.

That’s like having the next since you don’t have to pay that for.

Next, let’s say you have because you owe people .

Then somehow you spend another . Maybe you borrowed this amount.

So now, you owe . You actually have .

Here, you’re already negative but spent some more. So you are further negative.

Same thing with the third example.

You’re already at but spend another . So now you owe altogether.

You can also memorize this formula:

Here, we keep the first number then change the second sign and also change the third sign.

Just follow the formula: keep-change-change.

Keep the first number , then change the next sign from negative to positive and then change the sign of the last number from positive to negative .

Remember the rule for adding like signs? We just have to add the two numbers together and keep the sign.

Let’s do the third example.

Again, we’re going to do keep-change-change.

Keep the first one , change the negative sign to a positive and then change the positive sign to a negative in the last number.

Let me show you another example.

This isn’t too bad. Both numbers are postive. But we should expect the answer to be negative since we are subtracting more than what we have.

Let’s solve this using keep-change-change.

Keep the first number , change the next sign negative into positive and then change the positive sign of the last number to a negative sign.

Just to recap, when we subtract signed numbers, we can use keep-change-change to make it an addition problem and we’re just going to use our rules in adding signed numbers. This is how to add and subtract signed numbers.