Student Outcomes
Students know that a number raised to the zeroth power is equal to one.
Students recognize the need for the definition to preserve the properties of exponents.
Lesson Notes
In this lesson we introduce the zeroth power and its definition. Most of the time in this lesson should be spent having students work through possible meanings of numbers raised to the zeroth power and then checking the validity of their claims. For that reason, focus should be placed on the Exploratory Challenge in this lesson. Encourage students to share their thinking about what a number raised to the zeroth power could mean. It may be necessary to guide students through the work of developing cases to check the definition 𝑥𝑥0= 1 in Exercise 1 and the check of the first law of exponents in Exercise 2. Exercises 3 and 4 may be omitted if time is an issue.
Classwork
Concept Development (5 minutes): Let us summarize our main conclusions about exponents. For any numbers 𝑥𝑥, 𝑦𝑦 and any positive integers 𝑚𝑚, 𝑛𝑛, the following holds
𝑥𝑥𝑚𝑚∙ 𝑥𝑥𝑛𝑛=𝑥𝑥𝑚𝑚+𝑛𝑛 (1)
(𝑥𝑥𝑚𝑚)𝑛𝑛=𝑥𝑥𝑚𝑚𝑛𝑛 (2)
(𝑥𝑥𝑦𝑦)𝑛𝑛=𝑥𝑥𝑛𝑛𝑦𝑦𝑛𝑛. (3)
If we assume 𝑥𝑥 0 in equation (4) and 𝑦𝑦 0 in equation (5) below, then we also have
=𝑥𝑥𝑚𝑚−𝑛𝑛 (4)
�𝑥𝑥�𝑛𝑛 = 𝑥𝑥
𝑛𝑛
. (5)
We have shown that for any numbers , , and any positive integers , , the following holds
∙ = + (1)
( ) = (2)
( ) = . (3)
Definition: ___________________________________________________________________________
There is an obvious reason why the 𝑥𝑥 in (4) and the 𝑦𝑦 in (5) must be nonzero: We cannot divide by 0. Please note that in high school it is further necessary to restrict the values of 𝑥𝑥 and 𝑦𝑦 to nonnegative numbers when defining rational number exponents. Our examination of exponents in Grade 8 is limited to the integers, however, so restricting the base values to nonnegative numbers should not be a concern for students at this time.
We group equations (1)–(3) together because they are the foundation on which all the results about exponents rest.
When they are suitably generalized, as they are above, they imply (4) and (5). Therefore, we concentrate on (1)–(3).
The most important feature of (1)–(3) is that they are simple and formally (symbolically) natural. Mathematicians want these three identities to continue to hold for all exponents 𝑚𝑚 and 𝑛𝑛, without the restriction that 𝑚𝑚 and 𝑛𝑛 be positive integers because of these two desirable qualities. We should do it one step at a time. Our goal in this grade is to extend the validity of (1)–(3) to all integers 𝑚𝑚 and 𝑛𝑛.
Exploratory Challenge (20 minutes)
The first step in this direction is to introduce the definition of the 0th exponent of a number and to then use it to prove that (1)–(3) remain valid when 𝑚𝑚 and 𝑛𝑛 are not just positive integers but all whole numbers (including 0). Since our goal is to make sure (1)–(3) remain valid even when 𝑚𝑚 and 𝑛𝑛 may be 0, the very definition of the 0th exponent of a number must pose no obvious contradiction to (1)–(3). With this in mind, let us consider what it means to raise a number 𝑥𝑥 to the zeroth power. For example, what should 30 mean?
Students will likely respond that 30 should equal 0. When they do, demonstrate why that would contradict our existing understanding of properties of exponents using (1). Specifically, if 𝑚𝑚 is a positive integer and we let 30= 0, then
3𝑚𝑚∙30= 3𝑚𝑚+0,
but since we let 30= 0, it means that the left side of the equation would equal zero. That creates a contradiction because
0 3𝑚𝑚+0.
Therefore, letting 30= 0 will not help us to extend (1)–(3) to all whole numbers 𝑚𝑚 and 𝑛𝑛.
Next, students may say that we should let 30= 3. Show the two problematic issues this would create. First, in Lesson 1, we learned that, by definition, 𝑥𝑥1=𝑥𝑥, and we do not want to have two powers that yield the same result.
Second, it would violate the existing rules we have developed: Looking specifically at (1) again, if we let 30= 3, then
3𝑚𝑚∙30= 3𝑚𝑚+0, but
3𝑚𝑚∙30= 3 × × 3
𝑚𝑚 times
∙3
= 3𝑚𝑚+1, which again is a contradiction.
Scaffolding:
Ask struggling students to use what they know about the laws of exponents to rewrite 3𝑚𝑚 30. Their resulting expression should be 3𝑚𝑚. Then ask: If a number, 3, is multiplied by another number, and the product is 3, what does that mean about the other number? It means the other number must be 1. Then have students apply this thinking to the equation 3𝑚𝑚∙30= 3𝑚𝑚 to determine the value of 30.
If we believe that equation (1) should hold even when 𝑛𝑛= 0, then, for example, 32+0= 32× 30, which is the same as 32= 32× 30; therefore, after multiplying both sides by the number 1
3 , we get 1 = 30. In the same way, our belief that (1) should hold when either 𝑚𝑚 or 𝑛𝑛 is 0, would lead us to conclude that we should define 𝑥𝑥0= 1 for any nonzero 𝑥𝑥.
Therefore, we give the following definition:
Definition: For any positive number 𝑥𝑥, we define 𝑥𝑥0= 1.
Students should write this definition of 𝑥𝑥0 in the lesson summary box on their classwork paper.
Now that 𝑥𝑥𝑛𝑛 is defined for all whole numbers 𝑛𝑛, check carefully that (1)–(3) remain valid for all whole numbers 𝑚𝑚 and 𝑛𝑛.
Have students independently complete Exercise 1; provide correct values for 𝑚𝑚 and 𝑛𝑛 before proceeding to the development of cases (A)–(C).
Exercise 1
List all possible cases of whole numbers and for identity (1). More precisely, when > and > , we already know that (1) is correct. What are the other possible cases of and for which (1) is yet to be verified?
Case (A): > and = Case (B): = and >
Case (C): = =
Model how to check the validity of a statement using Case (A) with equation (1) as part of Exercise 2. Have students work independently or in pairs to check the validity of (1) in Case (B) and Case (C) to complete Exercise 2. Next, have students check the validity of equations (2) and (3) using Cases (A)–(C) for Exercises 3 and 4.
Exercise 2
Check that equation (1) is correct for each of the cases listed in Exercise 1.
Case (A): ∙ = ? Yes, because ∙ = ∙ = .
Case (B): ∙ = ? Yes, because ∙ = ∙ = .
Case (C): ∙ = ? Yes, because ∙ = ∙ = .
Exercise 3
Do the same with equation (2) by checking it case-by-case.
Case (A): ( ) = × ? Yes, because is a number, and a number raised to a zero power is . = = × . So, the left side is . The right side is also because × = = .
Case (B): ( ) = ×? Yes, because, by definition = and = , the left side is equal to . The right side is equal to = , so both sides are equal.
Case (C): ( ) = ×? Yes, because, by definition of the zeroth power of , both sides are equal to .
MP.6
MP.3
Exercise 4
Do the same with equation (3) by checking it case-by-case.
Case (A): ( ) = ? Yes, because the left side is by the definition of the zeroth power, while the right side is
× = .
Case (B): Since > , we already know that (3) is valid.
Case (C): This is the same as Case (A), which we have already shown to be valid.
Exploratory Challenge 2 (5 minutes)
Students practice writing numbers in expanded form in Exercises 5 and 6. Students use the definition of 𝑥𝑥0, for any number 𝑥𝑥, learned in this lesson.
Clearly state that you want to see the ones digit multiplied by 100. That is the important part of the expanded notation because it leads to the use of negative powers of 10 for decimals in Lesson 5.
Exercise 5
Write the expanded form of , using exponential notation.
= ( × ) + ( × ) + ( × ) + ( × )
Exercise 6
Write the expanded form of , , using exponential notation.
= ( × ) + ( × ) + ( × ) + ( × ) + ( × ) + ( × ) + ( × )
Closing (3 minutes)
Summarize, or have students summarize, the lesson.
The rules of exponents that we have worked on prior to today only work for positive integer exponents; now those same exponent rules have been extended to all whole numbers.
The next logical step is to attempt to extend these rules to all integer exponents.
Exit Ticket (2 minutes)
Fluency Exercise (10 minutes)
Sprint: Rewrite expressions with the same base for positive exponents only. Make sure to tell the students that all letters within the problems of the Sprint are meant to denote numbers. This exercise can be administered at any point during the lesson. Refer to the Sprints and Sprint Delivery Script sections in the Module Overview for directions to
Scaffolding:
You may need to remind students how to write numbers in expanded form with
Exercise 5.
Scaffolding:
Ask advanced learners to consider the expanded form of a number such as 945.78. Ask them if it is possible to write this number in expanded form using exponential notation. If so, what exponents would represent the decimal place values? This question serves as a sneak peek into the meaning of negative integer exponents.
Name ___________________________________________________ Date____________________
Lesson 4: Numbers Raised to the Zeroth Power
Exit Ticket
1. Simplify the following expression as much as possible.
410 410∙70=
2. Let 𝑎𝑎 and 𝑏𝑏 be two numbers. Use the distributive law and then the definition of zeroth power to show that the numbers (𝑎𝑎0+𝑏𝑏0)𝑎𝑎0 and (𝑎𝑎0+𝑏𝑏0)𝑏𝑏0 are equal.
Exit Ticket Sample Solutions
1. Simplify the following expression as much as possible.
∙ = − ∙ = ∙ = ∙ =
2. Let and be two numbers. Use the distributive law and then the definition of zeroth power to show that the numbers ( + ) and ( + ) are equal.
( + ) = ∙ + ∙
= + +
= +
= + ∙
= +
=
( + ) = ∙ + ∙
= + +
= +
= ∙ +
= +
= Since both numbers are equal to , they are equal.
Problem Set Sample Solutions
Let , be numbers ( , ). Simplify each of the following expressions.
1.
= −
= =
2.
∙ =
= −
=
=
3.
( ( . ) ) =
= ( . )×
= ( . )
=
4.
∙ ∙ ∙ = ∙
= − ∙ −
= ∙
= 5.
∙ = ∙
∙
= ∙
= − ∙ −
= ∙
=
Applying Properties of Exponents to Generate Equivalent Expressions—Round 1
Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. All letters denote numbers.
1. 22 ∙23 23. 63∙62
2. 22 ∙2 24. 62∙63
3. 22 ∙25 25. (−8)3∙(−8)
4. 3 ∙31 26. (−8) ∙(−8)3
5. 38 ∙31 27. (0.2)3∙(0.2)
6. 39∙31 28. (0.2) ∙(0.2)3
7. 76 ∙72 29. (−2)12∙(−2)1
8. 76 ∙73 30. (−2.7)12∙(−2.7)1
9. 76 ∙7 31. 1.16∙1.19
10. 1115∙11 32. 576∙579
11. 1116∙11 33. 𝑥𝑥6∙ 𝑥𝑥9
12. 212∙22 34. 2 ∙4
13. 212∙2 35. 2 ∙42
14. 212∙26 36. 2 ∙16
15. 995 ∙992 37. 16∙43
16. 996 ∙993 38. 32∙9
17. 99 ∙99 39. 32∙27
18. 58 ∙52 40. 32∙81
19. 68 ∙62 41. 5 ∙25
20. 78 ∙72 42. 5 ∙125
21. 8∙ 2 43. 8∙29
Number Correct: ______
Applying Properties of Exponents to Generate Equivalent Expressions—Round 1 [KEY]
Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. All letters denote numbers.
1. 22 ∙23 23. 63∙62
2. 22 ∙2 24. 62∙63
3. 22 ∙25 25. (−8)3∙(−8) (− )
4. 3 ∙31 26. (−8) ∙(−8)3 (− )
5. 38 ∙31 27. (0.2)3∙(0.2) ( . )
6. 39∙31 28. (0.2) ∙(0.2)3 ( . )
7. 76 ∙72 29. (−2)12∙(−2)1 (− )
8. 76 ∙73 30. (−2.7)12∙(−2.7)1 (− . )
9. 76 ∙7 31. 1.16∙1.19 .
10. 1115∙11 32. 576∙579
11. 1116∙11 33. 𝑥𝑥6∙ 𝑥𝑥9
12. 212∙22 34. 2 ∙4
13. 212∙2 35. 2 ∙42
14. 212∙26 36. 2 ∙16
15. 995 ∙992 37. 16∙43
16. 996 ∙993 38. 32∙9
17. 99 ∙99 39. 32∙27
18. 58 ∙52 40. 32∙81
19. 68 ∙62 41. 5 ∙25
20. 78 ∙72 42. 5 ∙125
21. 8∙ 2 43. 8∙29
Applying Properties of Exponents to Generate Equivalent Expressions—Round 2
Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. All letters denote numbers.
1. 52 ∙53 23. 73∙72
2. 52 ∙5 24. 72∙73
3. 52 ∙55 25. (−4)3∙(−4)11
4. 2 ∙21 26. (−4)11∙(−4)3
5. 28 ∙21 27. (0.2)3∙(0.2)11
6. 29∙21 28. (0.2)11∙(0.2)3
7. 36 ∙32 29. (−2)9∙(−2)5
8. 36 ∙33 30. (−2.7)5∙(−2.7)9
9. 36 ∙3 31. 3.16∙3.16
10. 715∙7 32. 576∙576
11. 716∙7 33. 6∙ 6
12. 1112∙112 34. 4∙29
13. 1112∙11 35. 42∙29
14. 1112∙116 36. 16∙29
15. 235 ∙232 37. 16∙43
16. 236 ∙233 38. 9∙35
17. 23 ∙23 39. 35∙9
18. 13 ∙133 40. 35∙27
19. 15 ∙153 41. 5 ∙25
20. 17 ∙173 42. 5 ∙125
21. 𝑥𝑥 ∙ 𝑥𝑥3 43. 211∙4
Improvement: ______
Applying Properties of Exponents to Generate Equivalent Expressions—Round 2 [KEY]
Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. All letters denote numbers.
1. 52 ∙53 23. 73∙72
2. 52 ∙5 24. 72∙73
3. 52 ∙55 25. (−4)3∙(−4)11 (− )
4. 2 ∙21 26. (−4)11∙(−4)3 (− )
5. 28 ∙21 27. (0.2)3∙(0.2)11 ( . )
6. 29∙21 28. (0.2)11∙(0.2)3 ( . )
7. 36 ∙32 29. (−2)9∙(−2)5 (− )
8. 36 ∙33 30. (−2.7)5∙(−2.7)9 (− . )
9. 36 ∙3 31. 3.16∙3.16 .
10. 715∙7 32. 576∙576
11. 716∙7 33. 6∙ 6
12. 1112∙112 34. 4∙29
13. 1112∙11 35. 42∙29
14. 1112∙116 36. 16∙29
15. 235 ∙232 37. 16∙43
16. 236 ∙233 38. 9∙35
17. 23 ∙23 39. 35∙9
18. 13 ∙133 40. 35∙27
19. 15 ∙153 41. 5 ∙25
20. 17 ∙173 42. 5 ∙125
21. 𝑥𝑥 ∙ 𝑥𝑥3 43. 211∙4