Lesson 21
Sums and Products of Rational and Irrational Numbers
21.1: Operations on Integers (5 minutes)
Warmup
This activity prompts students to recognize that the sum of any two integers is always an integer and that the product of any two integers is also always an integer. Students will not be justifying in a formal way as to why these properties are true. (In a future course, when studying polynomials, students will begin considering integers as a closed system under addition, subtraction, and multiplication.) Some students may simply assert that these are true because they are not able to find two integers that add up to or multiply to make a noninteger. Others may reason that:
 It seems impossible to add two whole amounts and end up with a partial amount in the sum.
 It seems impossible to multiply wholenumber groups of whole numbers to make a nonwhole number quantity.
 If we use the number line to represent addition of two integers, we will start at an integer and take wholenumber steps, so we cannot end up at a point between integer values.
These are valid conclusions at this stage. Later in the lesson, students will use these conclusions to reason about the sums and products of rational numbers.
Launch
Arrange students in groups of 2. Ask them to think quietly for a couple of minutes before discussing their thinking with a partner.
Student Facing
Here are some examples of integers:
 25
 10
 2
 1
 0
 5
 9
 40

Experiment with adding any two numbers from the list (or other integers of your choice). Try to find one or more examples of two integers that:
 add up to another integer
 add up to a number that is not an integer

Experiment with multiplying any two numbers from the list (or other integers of your choice). Try to find one or more examples of two integers that:
 multiply to make another integer
 multiply to make a number that is not an integer
Student Response
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Anticipated Misconceptions
Some students might persist in attempting to find examples of two integers that add up to a noninteger. Rather than giving them more time to find examples, encourage them to think about the placement of integers on a number line and what it might imply about the sum or product of any two integers.
Activity Synthesis
Select students or groups to share their examples and their challenges in finding examples for the second part of each question. Invite as many possible explanations as time permits as to why they could not find examples of two integers that add or multiply to make a number that is not an integer. If no students brought up reasons similar to those listed in the Activity Narrative, ask students to consider them.
Explain that while we have not proven that two integers can never produce a sum or a product that is not an integer, for now we will accept this to be true.
21.2: Sums and Products of Rational Numbers (15 minutes)
Activity
In this activity, students use logical reasoning to develop a general argument about why the sum and product of two rational numbers are also rational numbers.
Students begin by studying some numerical examples of addition of two rational numbers and articulating why the sums are rational numbers. Next, they reason more abstractly about rational numbers in the form of \(\frac{a}{b}\) and \(\frac{c}{d}\) (MP2). Students consolidate what they know about the sum and product of integers (that these are also integers) and about rational numbers (that these are fractions with integers in the numerator and in the denominator) to argue that the result of adding or multiplying any two rational numbers must be another rational number (MP3).
Launch
Remind students that rational numbers are fractions and their opposites. Ask students what kinds of numbers the numerator and denominator in a fraction are. (They are integers and the denominator is not 0.)
Display the four addition expressions in the first question for all to see. Ask students, “How do we know that the numbers being added are each a rational number?” (They can be written as a positive or negative fraction.)
Arrange students in groups of 2. Give students a moment to think quietly about the first question and then time to discuss their thinking with their partner. Pause for a class discussion before students proceed to the rest of the questions.
Make sure students understand how \(\frac 23+ \frac {a}{15}\) leads to \( \frac {10+a}{15}\) and can explain why this sum is a rational number. Remind students of a conclusion from the warmup: that the sum of two integers is always an integer. Because \(a\) is an integer, \(10+a\) must be an integer, and \(\frac {10+a}{15}\) is therefore a fraction.
Student Facing

Here are a few examples of adding two rational numbers. Is each sum a rational number? Be prepared to explain how you know.
 \(4 +0.175 = 4.175\)
 \(\frac12 + \frac45 = \frac {5}{10}+\frac{8}{10} = \frac{13}{10}\)
 \(\text0.75 + \frac{14}{8} = \frac {\text6}{8} + \frac {14}{8} = \frac 88 = 1\)
 \(a\) is an integer: \(\frac 23+ \frac {a}{15} =\frac{10}{15} + \frac {a}{15} = \frac {10+a}{15}\)

Here is a way to explain why the sum of two rational numbers is rational.
Suppose \(\frac{a}{b}\) and \(\frac{c}{d}\) are fractions. That means that \(a, b, c,\) and \(d\) are integers, and \(b\) and \(d\) are not 0.
 Find the sum of \(\frac{a}{b}\) and \(\frac{c}{d}\). Show your reasoning.
 In the sum, are the numerator and the denominator integers? How do you know?
 Use your responses to explain why the sum of \(\frac{a}{b} + \frac{c}{d}\) is a rational number.
 Use the same reasoning as in the previous question to explain why the product of two rational numbers, \(\frac{a}{b} \boldcdot \frac{c}{d}\), must be rational.
Student Response
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Student Facing
Are you ready for more?
Consider numbers that are of the form \(a + b \sqrt{5}\), where \(a\) and \(b\) are integers. Let’s call such numbers quintegers.
Here are some examples of quintegers:
 \(3 + 4\sqrt{5}\) (\(a=3\), \(b=4\))
 \(7  2\sqrt{5}\) (\(a=7\), \(b=\text2\))
 \(\text5 + \sqrt{5}\) (\(a=\text5\), \(b=1\))
 3 (\(a=3\), \(b=0\)).
 When we add two quintegers, will we always get another quinteger? Either prove this, or find two quintegers whose sum is not a quinteger.
 When we multiply two quintegers, will we always get another quinteger? Either prove this, or find two quintegers whose product is not a quinteger.
Student Response
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Anticipated Misconceptions
Some students may not recognize 0.175, 4.175, and 0.75 as rational numbers. Demonstrate that these numbers can be written as the fractions \(\frac{175}{1000}\), \(\frac{4175}{1000}\), and \(\text\frac{75}{100} \) (or \(\text\frac34\)).
Activity Synthesis
The second question guides students through the pieces needed to make an argument that the sum of two rational numbers must be rational. Use the discussion to help students consolidate these pieces into a logical and coherent argument:
 If \(\frac{a}{b}\) and \(\frac{c}{d}\) are fractions, then its sum is \(\frac {ad + bc}{bd}\). The expressions \(ad, bc\), and \(bd\) must be integers because they are each a product of integers, and consequently \(ad + bc\) must also be an integer. Because the numerator and the denominator are both integers, the number is a fraction and is therefore rational.
Make sure students see how to construct a similar argument for the product of two rational numbers, as shown in the student response.
Design Principle(s): Support sensemaking; Maximize metaawareness
21.3: Sums and Products of Rational and Irrational Numbers (15 minutes)
Activity
In the previous activity, students justified why the sum and product of two rational numbers are also rational. In this activity, they develop an argument to explain why the sum and product of a rational number and an irrational number are irrational.
Students encounter an argument by contradiction. They learn that we can first assume that the sum \(\text {rational} + \text {irrational}\) is rational. If this is indeed true, when this sum is added to another rational number, the result must also be rational. It turns out that this is not the case (adding this sum to a certain rational number produces the original irrational number), which means that the sum \(\text {rational} + \text {irrational}\) cannot be rational.
In the last part of the activity, students make a similar argument for why the product of a rational number and an irrational number is irrational.
Launch
Keep students in groups of 2. Ask students to think quietly about the first question before conferring with their partner.
Supports accessibility for: Language; Organization
Student Facing

Here is a way to explain why \(\sqrt2 + \frac 19\) is irrational.

Let \(s\) be the sum of \( \sqrt2\) and \(\frac 19\), or \(s=\sqrt2 + \frac 19\).

Suppose \(s\) is rational.
 Would \(s + \text \frac19\) be rational or irrational? Explain how you know.
 Evaluate \(s + \text\frac19\). Is the sum rational or irrational?
 Use your responses so far to explain why \(s\) cannot be a rational number, and therefore \( \sqrt2 + \frac 19\) cannot be rational.

 Use the same reasoning as in the earlier question to explain why \(\sqrt2 \boldcdot \frac 19\) is irrational.
Student Response
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Activity Synthesis
As in the previous activity, students are guided through the pieces needed to make a particular argument—that the sum of a rational number and an irrational number must be irrational. Make sure students can consolidate these pieces into a logical and coherent argument:
 Let \(s\) be the sum of \(\frac{a}{b}\) (rational) and \(\sqrt{p}\) (irrational).
 Suppose \(s\) is rational. Adding \(s\) and \(\text \frac{a}{b}\) must produce a rational number because the sum of two rational values is rational.
 Adding \(s\) and \(\text\frac{a}{b}\) also gives \(\frac{a}{b} + \sqrt{p} + \text \frac{a}{b}\) or \(\sqrt {p}\), which is irrational.
 The sum of \(s\) and \(\sqrt{p}\) cannot be both rational and irrational, so the assumption that \(s\) is rational must be false. So \(s\) (the sum of a rational number and an irrational number) must be irrational.
Make sure students see how to construct a similar argument for the product of a rational number and an irrational number, as shown in the Student Response.
Design Principle(s): Optimize output (for explanation); Cultivate conversation
21.4: Equations with Different Kinds of Solutions (30 minutes)
Optional activity
In this optional activity, students investigate how the parameters in a quadratic equation affect the number of solutions and the kinds of solutions. Students are first given the equation \(4x^2 + bx + 9=0\) and are asked to find a value of \(b\) that would produce a certain number or kind of solution. To do so, students may take different approaches, each with different degrees of efficiency.
Monitor for these strategies, from less efficient to more efficient:
 Substitute different integers for \(b\) and calculate the solutions by rewriting in factored form (if possible), completing the square, or using the quadratic formula.
 Enter the equation \(y=4x^2+bx+9\) in a graphing tool, graph the equation at different integer values of \(b\), and identify the \(x\)intercepts (and notice when there is only one or when there are none).
 Use a spreadsheet tool to calculate the solutions at various values of \(b\) (using the quadratic formula) and choose the values of \(b\) accordingly.
Students then study their findings and make general observations about which values of \(b\) produce 0, 1, or 2 solutions. Making generalization about the connections between \(b\) and the number of solutions encourages students to look for and make use of structure (MP7).
The last question is an openended task and may be challenging. It prompts students to write original equations that produce specified solutions. To do so effectively and efficiently, students need to make use of structure and any productive strategies seen in the first question and in past work (MP7) and to persevere (MP1). If time is limited and if desired, consider returning to this question at another time.
Depending on the strategy used, equations with rational solutions and those with single solutions might be harder to write than those with irrational solutions or no solutions. (The former requires making use of structure, while the latter could be achieved by choosing \(a, b\), and \(c\) somewhat randomly for \(ax^2 + bx+c=0\) and checking by graphing that the zeros appear irrational or are nonexistent.) If desired, differentiate the work by assigning certain questions to certain students or groups.
Making graphing and spreadsheet technology available gives students an opportunity to choose appropriate tools strategically (MP5).
Launch
Tell students that the numbers in a quadratic equation affect the type of solutions and the number of solutions. Display \(x^2 + bx + 4=0\) for all to see. Assign each student an integer between 8 and 8 (or another range that includes positive and negative values). Give students a moment to solve the equation with their assigned \(b\). Remind students that the solutions can be found by rewriting the equation in factored form, completing the square, using the quadratic formula, or graphing.
Ask students to report how many solutions they found at different values of \(b\) and what types of numbers the solutions are. Record and organize their findings in a table such as this one, displayed for all to see:
\(b\)  number of solutions  rational or irrational? 

8  two  irrational 
7  two  irrational 
...  
2  none  
1  none  
0  none  
1  none  
2  none  
3  none  
4  one  rational 
5  two  rational 
6  two  irrational 
Tell students that their job in this activity is to write and solve quadratic equations such that each equation has a particular kind or a particular number of solutions, and to think more generally about how the numbers in the equation relate to the solutions.
Arrange students in groups of 2–4. Encourage group members to collaborate and find different values of \(b\) in the first question so that they have a greater set of data to help them answer the second question. Ask them to consider helpful ways to collect and organize their findings and to think about efficient ways to find the right \(b\) values beyond trying different values of \(b\).
Student Facing

Consider the equation \(4x^2 + bx + 9=0\). Find a value of \(b\) so that the equation has:
 2 rational solutions
 2 irrational solutions
 1 solution
 no solutions
 Describe all the values of \(b\) that produce 2, 1, and no solutions.

Write a new quadratic equation with each type of solution. Be prepared to explain how you know that your equation has the specified type and number of solutions.
 no solutions
 2 irrational solutions
 2 rational solutions
 1 solution
Student Response
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Activity Synthesis
Select students or groups to share their strategies for finding the right values for \(b\) and for knowing which values of \(b\) would produce certain kinds or numbers of solutions.
Ask them to present in the order listed in the Activity Narrative. If only guessing and checking is mentioned by students, bring up at least one other strategy. Connect these strategies to what students learned about the equations and graphs of quadratic functions, for instance:
 The solutions to a quadratic equation \(ax^2+bx+c=0\) are represented by the \(x\)intercepts of the graph of \(y=ax^2+bx+c\). Graphing can be a handy way to see when an equation has one solution or no solution. It is less handy for telling if the solutions are rational or irrational.
 If a quadratic equation can be written as \(\text{factored form}=0\), we can tell that the solutions will be rational (provided that the factors do not contain irrational numbers).
 A quadratic equation that is of the form \(\text {perfect square}=0\) has two factors that are the same expression, which tells us that there is only one solution.
If time permits, consider demonstrating how graphing or spreadsheet technology could be used to help us spot patterns and suggest which values of \(a, b\), and \(c\) in a quadratic equation lead to certain solutions.
Design Principle(s): Cultivate conversation; Maximize metaawareness
Lesson Synthesis
Lesson Synthesis
Display a list of sums and products, such as shown:
 \(\frac {p}{q} + 9\), where \(p\) and \(q\) are integers and \(q\) is not zero. The sum is rational.
 \(\frac {p}{q} \boldcdot m\), where \(p\), \(q\), and \(m\) are all integers and \(q\) is not zero. The product is rational.
 \(5 \boldcdot \sqrt {11}\). The product is irrational.
 \(\frac98 + \sqrt {3}\). The sum is irrational.
Ask students to choose a sum or a product (or one of each). Then, ask them to practice explaining to 2–3 different partners why the sum or product must be rational or irrational, and to try making their explanation clearer and stronger each time.
21.5: Cooldown  Adding Irrational Numbers (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
We know that quadratic equations can have rational solutions or irrational solutions. For example, the solutions to \((x+3)(x1)=0\) are 3 and 1, which are rational. The solutions to \(x^28=0\) are \(\pm \sqrt{8}\), which are irrational.
Sometimes solutions to equations combine two numbers by addition or multiplication—for example, \(\pm 4\sqrt{3}\) and \(1 +\sqrt {12}\). What kind of number are these expressions?
When we add or multiply two rational numbers, is the result rational or irrational?

The sum of two rational numbers is rational. Here is one way to explain why it is true:
 Any two rational numbers can be written \(\frac{a}{b}\) and \(\frac{c}{d}\), where \(a, b, c, \text{ and } d\) are integers, and \(b\) and \(d\) are not zero.
 The sum of \(\frac{a}{b}\) and \(\frac{c}{d}\) is \(\frac{ad+bc}{bd}\). The denominator is not zero because neither \(b\) nor \(d\) is zero.
 Multiplying or adding two integers always gives an integer, so we know that \(ad, bc, bd\) and \(ad+bc\) are all integers.
 If the numerator and denominator of \(\frac{ad+bc}{bd}\) are integers, then the number is a fraction, which is rational.

The product of two rational numbers is rational. We can show why in a similar way:
 For any two rational numbers \(\frac{a}{b}\) and \(\frac{c}{d}\), where \(a, b, c, \text{ and } d\) are integers, and \(b\) and \(d\) are not zero, the product is \(\frac{ac}{bd}\).
 Multiplying two integers always results in an integer, so both \(ac\) and \(bd\) are integers, so \(\frac{ac}{bd}\) is a rational number.
What about two irrational numbers?

The sum of two irrational numbers could be either rational or irrational. We can show this through examples:
 \(\sqrt3\) and \(\text\sqrt3\) are each irrational, but their sum is 0, which is rational.
 \(\sqrt3\) and \(\sqrt5\) are each irrational, and their sum is irrational.

The product of two irrational numbers could be either rational or irrational. We can show this through examples:
 \(\sqrt2\) and \(\sqrt8\) are each irrational, but their product is \(\sqrt{16}\) or 4, which is rational.
 \(\sqrt2\) and \(\sqrt7\) are each irrational, and their product is \(\sqrt{14}\), which is not a perfect square and is therefore irrational.
What about a rational number and an irrational number?

The sum of a rational number and an irrational number is irrational. To explain why requires a slightly different argument:
 Let \(R\) be a rational number and \(I\) an irrational number. We want to show that \(R+I\) is irrational.
 Suppose \(s\) represents the sum of \(R\) and \(I\) (\(s=R+I\)) and suppose \(s\) is rational.
 If \(s\) is rational, then \(s + \textR\) would also be rational, because the sum of two rational numbers is rational.
 \(s + \textR\) is not rational, however, because \((R + I) + \textR = I\).
 \(s + \textR\) cannot be both rational and irrational, which means that our original assumption that \(s\) was rational was incorrect. \(s\), which is the sum of a rational number and an irrational number, must be irrational.

The product of a nonzero rational number and an irrational number is irrational. We can show why this is true in a similar way:
 Let \(R\) be rational and \(I\) irrational. We want to show that \(R \boldcdot I\) is irrational.
 Suppose \(p\) is the product of \(R\) and \(I\) (\(p=R \boldcdot I\)) and suppose \(p\) is rational.
 If \(p\) is rational, then \(p \boldcdot \frac{1}{R}\) would also be rational because the product of two rational numbers is rational.
 \(p \boldcdot \frac{1}{R}\) is not rational, however, because \(R \boldcdot I \boldcdot \frac{1}{R} = I\).
 \(p \boldcdot \frac{1}{R}\) cannot be both rational and irrational, which means our original assumption that \(p\) was rational was false. \(p\), which is the product of a rational number and an irrational number, must be irrational.