5.3.2: Multiplying Rational Numbers - Mathematics

$5.3.2: Multiplying Rational Numbers - Mathematics$

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Lesson

Let's multiply signed numbers.

Exercise (PageIndex{1}): Before and After

Where was the girl:

5 seconds after this picture was taken? Mark her approximate location on the picture.
5 seconds before this picture was taken? Mark her approximate location on the picture.

Exercise (PageIndex{2}): Backwards in Time

A traffic safety engineer was studying travel patterns along a highway. She set up a camera and recorded the speed and direction of cars and trucks that passed by the camera. Positions to the east of the camera are positive, and to the west are negative.

Here are some positions and times for one car:
position (feet) (-180) (-120) (-60) (0) (60) (120)
time (seconds) (-3) (-2) (-1) (0) (1) (2)
Table (PageIndex{1})
1. In what direction is this car traveling?
2. What is its velocity?
1. What does it mean when the time is zero?
2. What could it mean to have a negative time?
Here are the positions and times for a different car whose velocity is -50 feet per second:
position (feet) (0) (-50) (-100)
time (seconds) (-3) (-2) (-1) (0) (1) (2)
Table (PageIndex{2})
1. Complete the table with the rest of the positions.
2. In what direction is this car traveling? Explain how you know.

position (feet)	(-180)	(-120)	(-60)	(0)	(60)	(120)
time (seconds)	(-3)	(-2)	(-1)	(0)	(1)	(2)

position (feet)	(0)	(-50)	(-100)
time (seconds)	(-3)	(-2)	(-1)	(0)	(1)	(2)

Complete the table for several different cars passing the camera.

Table (PageIndex{3})
velocity (meters per second)	time after passing the camera (seconds)	ending position (meters)	equation
car C	(+25)	(+10)	(+250)	(25cdot 10=250)
car D	(-20)	(+30)
car E	(+32)	(-40)
car F	(-35)	(-20)
car G	(-15)	(-8)

1. If a car is traveling east when it passes the camera, will its position be positive or negative 60 seconds before it passes the camera?
2. If we multiply a positive number and a negative number, is the result positive or negative?
1. If a car is traveling west when it passes the camera, will its position be positive or negative 60 seconds before it passes the camera?
2. If we multiply two negative numbers, is the result positive or negative?

Exercise (PageIndex{3}): Cruising

Around noon, a car was traveling -32 meters per second down a highway. At exactly noon (when time was 0), the position of the car was 0 meters.

Complete the table.
time (s) (-10) (-7) (-4) (-1) (2) (5) (8) (11)
position (m)
Table (PageIndex{4})
Graph the relationship between the time and the car's position.
1. What was the position of the car at -3 seconds?
2. What was the position of the car at 6.5 seconds?

Table (PageIndex{4})
time (s)	(-10)	(-7)	(-4)	(-1)	(2)	(5)	(8)	(11)
position (m)

Are you ready for more?

Find the value of these expressions without using a calculator.

((-1)^{2}qquad (-1)^{3}qquad (-1)^{4}qquad (-1)^{99})

Exercise (PageIndex{4}): Rational Numbers Multiplication Grid

Look at the patterns along the rows and columns and continue those patterns to complete the table. When you have filled in all the boxes you can see, click on the "More Boxes" button.

What does this tell you about multiplication by a negative?

Summary

We can use signed numbers to represent time relative to a chosen point in time. We can think of this as starting a stopwatch. The positive times are after the watch starts, and negative times are times before the watch starts.

If a car is at position 0 and is moving in a positive direction, then for times after that (positive times), it will have a positive position. A positive times a positive is positive.

If a car is at position 0 and is moving in a negative direction, then for times after that (positive times), it will have a negative position. A negative times a positive is negative.

If a car is at position 0 and is moving in a positive direction, then for times before that (negative times), it must have had a negative position. A positive times a negative is negative.

If a car is at position 0 and is moving in a negative direction, then for times before that (negative times), it must have had a positive position. A negative times a negative is positive.

Here is another way of seeing this:

We can think of (3cdot 5) as (5+5+5), which has a value of 15.

We can think of (3cdot (-5)) as (-5+-5+-5), which has a value of -15.

We know we can multiply positive numbers in any order: (3cdot 5=5cdot 3)

If we can multiply signed numbers in any order, then ((-5)cdot 3) would also equal -15.

Now let’s think about multiplying two negatives.

We can find (-5cdot (3+-3)) in two ways:

Applying the distributive property:
(-5cdot 3+-5cdot (-3))
Adding the numbers in parentheses:
(-5cdot (0)=0)

This means that these expressions must be equal.

(-5cdot 3+-5cdot (-3)=0)

Multiplying the first two numbers gives

(-15+-5cdot (-3)=0)

Which means that

(-5cdot (-3)=15)

There was nothing special about these particular numbers. This always works!

A positive times a positive is always positive.
A negative times a positive or a positive times a negative is always negative.
A negative times a negative is always positive.

Practice

Exercise (PageIndex{5})

Fill in the missing numbers in these equations

(-2cdot (-4.5)=?)
((-8.7)cdot (-10)=?)
((-7)cdot ?=14)
(?cdot (-10)=90)

Exercise (PageIndex{6})

A weather station on the top of a mountain reports that the temperature is currently (0^{circ} ext{C}) and has been falling at a constant rate of (3^{circ} ext{C}) per hour. If it continues to fall at this rate, find each indicated temperature. Explain or show your reasoning.

What will the temperature be in 2 hours?
What will the temperature be in 5 hours?
What will the temperature be in half an hour?
What was the temperature 1 hour ago?
What was the temperature 3 hours ago?
What was the temperature 4.5 hours ago?

Exercise (PageIndex{7})

Find the value of each expression.

(frac{1}{4}cdot (-12))
(-frac{1}{3}cdot 39)
((-frac{4}{5})cdot (-75))
(-frac{2}{5}cdot (-frac{3}{4}))
(frac{8}{3}cdot -42)

Exercise (PageIndex{8})

To make a specific hair dye, a hair stylist uses a ratio of (1frac{1}{8}) oz of red tone, (frac{3}{4}) oz of gray tone, and (frac{5}{8}) oz of brown tone.

If the stylist needs to make 20 oz of dye, how much of each dye color is needed?
If the stylist needs to make 100 oz of dye, how much of each dye color is needed?

(From Unit 4.1.2)

Exercise (PageIndex{9})

Here are the vertices of rectangle (FROG): ((-2,5),: (-2,1),: (6,5),: (6,1)).
Find the perimeter of this rectangle. If you get stuck, try plotting the points on a coordinate plane.
Find the area of the rectangle (FROG).
Here are the coordinates of rectangle (PLAY): ((-11,20),: (-11,-3),: (-1,20),: (-1,-3)). Find the perimeter and area of this rectangle. See if you can figure out its side lengths without plotting the points.

(From Unit 5.2.6)

Multiply Rational Numbers

Related Topics:
Common Core for Grade 7
Common Core for Mathematics
Lesson Plans and Worksheets for all Grades
More Lessons for Grade 7

Examples, solutions, worksheets, videos, and lessons to help Grade 7 students learn how to apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.

Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as (–1)(–1) = 1 and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts.

Suggested Learning Targets

I can multiply and divide rational numbers (integers, fractions, and decimals).
I can use the multiplication rules for integers and apply them to multiplying decimals and fractions.
I can use real-world contexts to describe the product of rational numbers.
I can iInterpret products of rational numbers in real world contexts.
I can create an equivalent mathematical expression when given an expression by using the distributive property or other properties of operations.
I can identify equivalent expressions when given two or more expressions.

Multiplying Rational Numbers
How to multiply rational numbers like decimals and fractions?

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.

Multiplying Rational Numbers Digital Math Activity

This digital math activity allows students to practice multiplying rational numbers, including word problems. The activity includes 4 interactive slides (ex: drag and match, using the typing tool, using the shape tool) and is paperless through Google Slides™ and PowerPoint™.

Math Concept:

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2. Exit Tickets (DIGITAL)

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3. Video and Step-by-Step Instructions

A teacher guide with both video and step-by-step instructions have been included to help you implement these activities into your classroom.

Please download the preview to see more pictures.

** The only portion of this that is editable is the Google Form™ exit ticket.**

Looking for additional 7th Grade Digital Math Activities? See our digital activity sets:

What do I need to use these files? You and your students will need access to Google Slides™ and Google Forms™.

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Maneuvering the Middle Resources

Maneuvering the Middle online for teacher tips, tricks, and math content!

Maneuvering the Middle Terms of Use

Products by Maneuvering the Middle, LLC may be used by the purchaser for their classroom use only. This is a single classroom license only. All rights reserved. Resources may only be posted online in an LMS such as Google Classroom, Canvas, or Schoology. Students should be the only ones able to access the resources. It is a copyright violation to upload the files to school/district servers or shared Google Drives. See more information on our terms of use here.

Important facts about rational numbers worksheets for Grade 6

Understanding the concept of rational numbers may seem complex, but our worksheets are here to inspire kids on very simple strategies to solve these operations. Moreover, our amazing worksheets will not only help your kids to understand this concept in a better way, but will equally make clear the distinction between rational and irrational numbers.

How does rational numbers fit within the real number system?

In a bid to understand how rational numbers fit within the real number system, we must first of all recall that, all rational numbers are real numbers, but not all real numbers are rational numbers. Real numbers are either rational or irrational.

For instance, decimal numbers with repeated decimals like 0.7373737373 are rational numbers while numbers like 0.2810582107432, whose decimals are not repeated, are irrational.

Secondly, unlike the rational number &radic9, that can be simplified, &radic7 is an irrational number, because it cannot be simplified.

In addition, a fraction with a denominator of zero is an irrational number (5/0). For a fraction to be considered a rational number, its numerator and denominator must be integers.

Secret to understanding rational numbers

Refresh your kid&rsquos basic math skills with these stimulating operations with rational numbers worksheets. In fact, the only secret to understanding operations with rational numbers requires just a consistent engagement in our rational numbers worksheets for grade 6 pdf.

However, given our many fun and challenging exercises on operations with rational numbers, your 6 th graders will obtain the best of absolute value skills, addition, and subtraction, multiplication and division skills.

5.3.2: Multiplying Rational Numbers - Mathematics

We now need to look at rational expressions. A rational expression is nothing more than a fraction in which the numerator and/or the denominator are polynomials. Here are some examples of rational expressions.

The last one may look a little strange since it is more commonly written (4 + 6x - 10). However, it’s important to note that polynomials can be thought of as rational expressions if we need to, although they rarely are.

There is an unspoken rule when dealing with rational expressions that we now need to address. When dealing with numbers we know that division by zero is not allowed. Well the same is true for rational expressions. So, when dealing with rational expressions we will always assume that whatever (x) is it won’t give division by zero. We rarely write these restrictions down, but we will always need to keep them in mind.

For the first one listed we need to avoid (x = 1). The second rational expression is never zero in the denominator and so we don’t need to worry about any restrictions. Note as well that the numerator of the second rational expression will be zero. That is okay, we just need to avoid division by zero. For the third rational expression we will need to avoid (m = 3) and (m = - 2). The final rational expression listed above will never be zero in the denominator so again we don’t need to have any restrictions.

The first topic that we need to discuss here is reducing a rational expression to lowest terms. A rational expression has been reduced to lowest terms if all common factors from the numerator and denominator have been canceled. We already know how to do this with number fractions so let’s take a quick look at an example.

With rational expression it works exactly the same way.

We do have to be careful with canceling however. There are some common mistakes that students often make with these problems. Recall that in order to cancel a factor it must multiply the whole numerator and the whole denominator. So, the x+3 above could cancel since it multiplied the whole numerator and the whole denominator. However, the (x)’s in the reduced form can’t cancel since the (x) in the numerator is not times the whole numerator.

To see why the (x)’s don’t cancel in the reduced form above put a number in and see what happens. Let’s plug in (x = 4).

Clearly the two aren’t the same number!

So, be careful with canceling. As a general rule of thumb remember that you can’t cancel something if it’s got a “+” or a “-” on one side of it. There is one exception to this rule of thumb with “-” that we’ll deal with in an example later on down the road.

Let’s take a look at a couple of examples.

(displaystyle frac <<- 2x - 8>><<- 9x + 20>>)
(displaystyle frac <<- 25>><<5x - >>)
(displaystyle frac <<+ 2 + >><<< ight)>^8>>>)

When reducing a rational expression to lowest terms the first thing that we will do is factor both the numerator and denominator as much as possible. That should always be the first step in these problems.

Also, the factoring in this section, and all successive section for that matter, will be done without explanation. It will be assumed that you are capable of doing and/or checking the factoring on your own. In other words, make sure that you can factor!

We’ll first factor things out as completely as possible. Remember that we can’t cancel anything at this point in time since every term has a “+” or a “-” on one side of it! We’ve got to factor first!

At this point we can see that we’ve got a common factor in both the numerator and the denominator and so we can cancel the (x)-4 from both. Doing this gives,

This is also all the farther that we can go. Nothing else will cancel and so we have reduced this expression to lowest terms.

Again, the first thing that we’ll do here is factor the numerator and denominator.

At first glance it looks there is nothing that will cancel. Notice however that there is a term in the denominator that is almost the same as a term in the numerator except all the signs are the opposite.

We can use the following fact on the second term in the denominator.

This is commonly referred to as factoring a minus sign out because that is exactly what we’ve done. There are two forms here that cover both possibilities that we are liable to run into. In our case however we need the first form.

Because of some notation issues let’s just work with the denominator for a while.

Notice the steps used here. In the first step we factored out the minus sign, but we are still multiplying the terms and so we put in an added set of brackets to make sure that we didn’t forget that. In the second step we acknowledged that a minus sign in front is the same as multiplication by “-1”. Once we did that we didn’t really need the extra set of brackets anymore so we dropped them in the third step. Next, we recalled that we change the order of a multiplication if we need to so we flipped the (x) and the “-1”. Finally, we dropped the “-1” and just went back to a negative sign in the front.

Typically, when we factor out minus signs we skip all the intermediate steps and go straight to the final step.

Let’s now get back to the problem. The rational expression becomes,

At this point we can see that we do have a common factor and so we can cancel the x-5.

In this case the denominator is already factored for us to make our life easier. All we need to do is factor the numerator.

Now we reach the point of this part of the example. There are 5 (x)’s in the numerator and 3 in the denominator so when we cancel there will be 2 left in the numerator. Likewise, there are 2 (left( ight))’s in the numerator and 8 in the denominator so when we cancel there will be 6 left in the denominator. Here is the rational expression reduced to lowest terms.

Before moving on let’s briefly discuss the answer in the second part of this example. Notice that we moved the minus sign from the denominator to the front of the rational expression in the final form. This can always be done when we need to. Recall that the following are all equivalent.

In other words, a minus sign in front of a rational expression can be moved onto the whole numerator or whole denominator if it is convenient to do that. We do have to be careful with this however. Consider the following rational expression.

In this case the “-” on the (x) can’t be moved to the front of the rational expression since it is only on the (x). In order to move a minus sign to the front of a rational expression it needs to be times the whole numerator or denominator. So, if we factor a minus out of the numerator we could then move it into the front of the rational expression as follows,

The moral here is that we need to be careful with moving minus signs around in rational expressions.

We now need to move into adding, subtracting, multiplying and dividing rational expressions.

Let’s start with multiplying and dividing rational expressions. The general formulas are as follows,

Note the two different forms for denoting division. We will use either as needed so make sure you are familiar with both. Note as well that to do division of rational expressions all that we need to do is multiply the numerator by the reciprocal of the denominator (i.e. the fraction with the numerator and denominator switched).

Before doing a couple of examples there are a couple of special cases of division that we should look at. In the general case above both the numerator and the denominator of the rational expression are fractions, however, what if one of them isn’t a fraction. So let’s look at the following cases.

Students often make mistakes with these initially. To correctly deal with these we will turn the numerator (first case) or denominator (second case) into a fraction and then do the general division on them.

Be careful with these cases. It is easy to make a mistake with these and incorrectly do the division.

Now let’s take a look at a couple of examples.

(displaystyle frac <<- 5x - 14>><<- 3x + 2>>,centerdot ,frac <<- 4>><<- 14x + 49>>)
(displaystyle frac <<- 9>><<+ 5m + 6>> div frac<<3 - m>><>)
(displaystyle frac <<+ 5y + 4>><- 1>><>>>)

Notice that with this problem we have started to move away from (x) as the main variable in the examples. Do not get so used to seeing (x)’s that you always expect them. The problems will work the same way regardless of the letter we use for the variable so don’t get excited about the different letters here.

Okay, this is a multiplication. The first thing that we should always do in the multiplication is to factor everything in sight as much as possible.

Now, recall that we can cancel things across a multiplication as follows.

Note that this ONLY works for multiplication and NOT for division!

In this case we do have multiplication so cancel as much as we can and then do the multiplication to get the answer.

With division problems it is very easy to mistakenly cancel something that shouldn’t be canceled and so the first thing we do here (before factoring. ) is do the division. Once we’ve done the division we have a multiplication problem and we factor as much as possible, cancel everything that can be canceled and finally do the multiplication.

So, let’s get started on this problem.

Now, notice that there will be a lot of canceling here. Also notice that if we factor a minus sign out of the denominator of the second rational expression. Let’s do some of the canceling and then do the multiplication.

Remember that when we cancel all the terms out of a numerator or denominator there is actually a “1” left over! Now, we didn’t finish the canceling to make a point. Recall that at the start of this discussion we said that as a rule of thumb we can only cancel terms if there isn’t a “+” or a “-” on either side of it with one exception for the “-”. We are now at that exception. If there is a “-” in front of the whole numerator or denominator, as we’ve got here, then we can still cancel the term. In this case the “-” acts as a “-1” that is multiplied by the whole denominator and so is a factor instead of an addition or subtraction. Here is the final answer for this part.

In this case all the terms canceled out and we were left with a number. This doesn’t happen all that often, but as this example has shown it clearly can happen every once in a while so don’t get excited about it when it does happen.

This is one of the special cases for division. So, as with the previous part, we will first do the division and then we will factor and cancel as much as we can.

Here is the work for this part.

Okay, it’s time to move on to addition and subtraction of rational expressions. Here are the general formulas.

As these have shown we’ve got to remember that in order to add or subtract rational expression or fractions we MUST have common denominators. If we don’t have common denominators then we need to first get common denominators.

Let’s remember how do to do this with a quick number example.

In this case we need a common denominator and recall that it’s usually best to use the least common denominator, often denoted lcd. In this case the least common denominator is 12. So we need to get the denominators of these two fractions to a 12. This is easy to do. In the first case we need to multiply the denominator by 2 to get 12 so we will multiply the numerator and denominator of the first fraction by 2. Remember that we’ve got to multiply both the numerator and denominator by the same number since we aren’t allowed to actually change the problem and this is equivalent to multiplying the fraction by 1 since (frac = 1). For the second term we’ll need to multiply the numerator and denominator by a 3.

Now, the process for rational expressions is identical. The main difficulty is in finding the least common denominator. However, there is a really simple process for finding the least common denominator for rational expressions. Here is it.

(displaystyle frac<4><<6>> - frac<1><<3>> + frac<5><<2>>)
(displaystyle frac<2><> - frac<><>)
(displaystyle frac<<- 2y + 1>> - frac<2><> + frac<3><>)
(displaystyle frac<<2x>><<- 9>> - frac<1><> - frac<2><>)
(displaystyle frac<4><> - frac<1>+ 1)

For this problem there are coefficients on each term in the denominator so we’ll first need the least common denominator for the coefficients. This is 6. Now, (x) (by itself with a power of 1) is the only factor that occurs in any of the denominators. So, the least common denominator for this part is (x) with the largest power that occurs on all the (x)’s in the problem, which is 5. So, the least common denominator for this set of rational expression is

So, we simply need to multiply each term by an appropriate quantity to get this in the denominator and then do the addition and subtraction. Let’s do that.

In this case there are only two factors and they both occur to the first power and so the least common denominator is.

Now, in determining what to multiply each part by simply compare the current denominator to the least common denominator and multiply top and bottom by whatever is “missing”. In the first term we’re “missing” a (z + 2) and so that’s what we multiply the numerator and denominator by. In the second term we’re “missing” a (z + 1) and so that’s what we’ll multiply in that term.

Here is the work for this problem.

The final step is to do any multiplication in the numerator and simplify that up as much as possible.

Be careful with minus signs and parenthesis when doing the subtraction.

Let’s first factor the denominators and determine the least common denominator.

So, there are two factors in the denominators a y-1 and a y+2. So we will write both of those down and then take the highest power for each. That means a 2 for the y-1 and a 1 for the y+2. Here is the least common denominator for this rational expression.

Now determine what’s missing in the denominator for each term, multiply the numerator and denominator by that and then finally do the subtraction and addition.

Okay now let’s multiply the numerator out and simplify. In the last term recall that we need to do the multiplication prior to distributing the 3 through the parenthesis. Here is the simplification work for this part.

Again, factor the denominators and get the least common denominator.

The least common denominator is,

Notice that the first rational expression already contains this in its denominator, but that is okay. In fact, because of that the work will be slightly easier in this case. Here is the subtraction for this problem.

Notice that we can actually go one step further here. If we factor a minus out of the numerator we can do some canceling.

Sometimes this kind of canceling will happen after the addition/subtraction so be on the lookout for it.

The point of this problem is that “1” sitting out behind everything. That isn’t really the problem that it appears to be. Let’s first rewrite things a little here.

In this way we see that we really have three fractions here. One of them simply has a denominator of one. The least common denominator for this part is,

Here is the addition and subtraction for this problem.

Notice the set of parenthesis we added onto the second numerator as we did the subtraction. We are subtracting off the whole numerator and so we need the parenthesis there to make sure we don’t make any mistakes with the subtraction.

Solved Examples

Find the difference between (egin frac <-5> <7>end) and (egin frac <3> <7>end)

The given rational numbers have a common denominator.

Subtract the numerators and retain the same denominator.

( herefore) The difference is (eginfrac <-8><7>end)

Example 2

Henry was shown the figure below.

How can he prove that the sum of the left-hand side figures is equal to the right-hand side?

The above two left-hand side figures will be mathematically represented as,

Let's follow the three steps to solve this problem.

They are the same in the above equation.

Since the denominators are the same, the numerators can be added and placed over the same denominator.

The fraction ( dfrac<5> <8>) is already in its simplest form.

The left-hand side picture shows the rational number ( dfrac<5> <8>) which is the same as that on the right-hand side.

( herefore) The sum on both the sides in the figure are the same.

Example 3

Sara uses (dfrac<3><5>) of the flour if she has to bake a full cake.

How much flour will she use to bake (dfrac<1><6>) portion of the cake?

She uses (dfrac<3><5>) of the flour to bake a full cake.

In (dfrac<1><6>) portion of the cake, she will use

( herefore) She would have to use (dfrac<1><10>) of the flour.

When (dfrac<6><7>y) is subtracted from a certain number, we get (dfrac<-14><8>y) as the answer. Find the number.
Solve: ((dfrac<-2><3>+dfrac<4><5>) imes dfrac<5><6>)

Lesson 9

The purpose of this lesson is to develop the rules for multiplying two negative numbers. Students use the familiar fact that (displaystyle mbox = mbox imesmbox) to make sense of this rule. They interpret negative time as time before a chosen starting time and then figure out what the position is of an object moving with a negative velocity at a negative time. An object moving with a negative velocity is moving from right to left along the number line. At a negative time it has not yet reached its starting point of zero, so it is to the right of zero, and therefore its position is positive. So a negative velocity times a negative time gives a positive position. When students connect reasoning about quantities with abstract properties of numbers, they engage in MP2.

There is also an optional activity where students can use another approach to understanding why the product of two negative numbers is positive, by examining patterns in an extended multiplication table that includes both positive and negative numbers (MP7).

Learning Goals

Let's multiply signed numbers.

Required Materials

Required Preparation

It is optional to provide 1 copy of the Rational Numbers Multiplication Grid blackline master to each student.

Recall that the notation a m means to multiply m instances of a.
Example: 2 4 = 2 &sdot 2 &sdot 2 &sdot 2

The expression ( − a ) m means to multiply m instances of the opposite of a (that is, m instances of &ndasha ). The notation − a m means to multiply m instances of a and then find the opposite of the result.

Find the value of each expression.
- − 2 4 =
- ( − 2 ) 3 =
- − ( − 7 ) 2 =
- 5 3 =
- What happens to the products as the first factor decreases by 1?
- How does the product compare to the product you would get if the factor was positive?
If you want $wv = r$ where $w,v$ are irrational and $r$ is rational, that must mean that $w =frac rv$ (since $v$ is irrational it isn't $ so we don't have to worry about that)

So to find any pair you want just pick any arbitrary rational number (so long as it is not zero as $w e 0$ and $v e 0$ then $wv e 0$ but that is the only restriction we can pick any other rational at all), and pick any arbitrary irrational $v$ . Then let $w = frac rv$ . As $v$ is irrational and $r$ is non-zero rational we will have $w = frac rv$ be irrational.

5.3.2: Multiplying Rational Numbers - Mathematics

Now that we have looked at integer exponents we need to start looking at more complicated exponents. In this section we are going to be looking at rational exponents. That is exponents in the form

where both (m) and (n) are integers.

We will start simple by looking at the following special case,

where (n) is an integer. Once we have this figured out the more general case given above will actually be pretty easy to deal with.

Let’s first define just what we mean by exponents of this form.

In other words, when evaluating (>>) we are really asking what number (in this case (a)) did we raise to the (n) to get (b). Often (>>) is called the (n) th root of b.

Let’s do a couple of evaluations.

When doing these evaluations, we will not actually do them directly. When first confronted with these kinds of evaluations doing them directly is often very difficult. In order to evaluate these we will remember the equivalence given in the definition and use that instead.

We will work the first one in detail and then not put as much detail into the rest of the problems.

So, here is what we are asking in this problem.

Using the equivalence from the definition we can rewrite this as,

So, all that we are really asking here is what number did we square to get 25. In this case that is (hopefully) easy to get. We square 5 to get 25. Therefore,

So what we are asking here is what number did we raise to the 5 th power to get 32?

What number did we raise to the 4 th power to get 81?

We need to be a little careful with minus signs here, but other than that it works the same way as the previous parts. What number did we raise to the 3 rd power (i.e. cube) to get -8?

This part does not have an answer. It is here to make a point. In this case we are asking what number do we raise to the 4 th power to get -16. However, we also know that raising any number (positive or negative) to an even power will be positive. In other words, there is no real number that we can raise to the 4 th power to get -16.

Note that this is different from the previous part. If we raise a negative number to an odd power we will get a negative number so we could do the evaluation in the previous part.

As this part has shown, we can’t always do these evaluations.

Again, this part is here to make a point more than anything. Unlike the previous part this one has an answer. Recall from the previous section that if there aren’t any parentheses then only the part immediately to the left of the exponent gets the exponent. So, this part is really asking us to evaluate the following term.

So, we need to determine what number raised to the 4 th power will give us 16. This is 2 and so in this case the answer is,

As the last two parts of the previous example has once again shown, we really need to be careful with parenthesis. In this case parenthesis makes the difference between being able to get an answer or not.

Also, don’t be worried if you didn’t know some of these powers off the top of your head. They are usually fairly simple to determine if you don’t know them right away. For instance, in the part b we needed to determine what number raised to the 5 will give 32. If you can’t see the power right off the top of your head simply start taking powers until you find the correct one. In other words compute (<2^5>), (<3^5>), (<4^5>) until you reach the correct value. Of course, in this case we wouldn’t need to go past the first computation.

The next thing that we should acknowledge is that all of the properties for exponents that we gave in the previous section are still valid for all rational exponents. This includes the more general rational exponent that we haven’t looked at yet.

Now that we know that the properties are still valid we can see how to deal with the more general rational exponent. There are in fact two different ways of dealing with them as we’ll see. Both methods involve using property 2 from the previous section. For reference purposes this property is,

So, let’s see how to deal with a general rational exponent. We will first rewrite the exponent as follows.

In other words, we can think of the exponent as a product of two numbers. Now we will use the exponent property shown above. However, we will be using it in the opposite direction than what we did in the previous section. Also, there are two ways to do it. Here they are,

Using either of these forms we can now evaluate some more complicated expressions

We can use either form to do the evaluations. However, it is usually more convenient to use the first form as we will see.

Let’s use both forms here since neither one is too bad in this case. Let’s take a look at the first form.

Now, let’s take a look at the second form.

So, we get the same answer regardless of the form. Notice however that when we used the second form we ended up taking the 3 rd root of a much larger number which can cause problems on occasion.

Again, let’s use both forms to compute this one.

As this part has shown the second form can be quite difficult to use in computations. The root in this case was not an obvious root and not particularly easy to get if you didn’t know it right off the top of your head.

In this case we’ll only use the first form. However, before doing that we’ll need to first use property 5 of our exponent properties to get the exponent onto the numerator and denominator.

We can also do some of the simplification type problems with rational exponents that we saw in the previous section.

For this problem we will first move the exponent into the parenthesis then we will eliminate the negative exponent as we did in the previous section. We will then move the term to the denominator and drop the minus sign.

In this case we will first simplify the expression inside the parenthesis.

Don’t worry if, after simplification, we don’t have a fraction anymore. That will happen on occasion. Now we will eliminate the negative in the exponent using property 7 and then we’ll use property 4 to finish the problem up.

We will leave this section with a warning about a common mistake that students make in regard to negative exponents and rational exponents. Be careful not to confuse the two as they are totally separate topics.

Watch the video: Ch Multiplying Negative Numbers (May 2022).