Adventures in Exploding Dots

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Adventures in

Exploding Dots

Lesson 8 PRACTICE PROBLEMS

Irrational Numbers and Infinite Sums

Playing with infinitely long sums and infinitely long polynomials to discover

irrational numbers, the geometric series formula, and more!

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We provide here a sampler of grade-appropriate—and optional!—practice questions you and your students might enjoy pursuing after lesson 8.

Feel free to cut-and-paste and/or adapt and/or use the questions here as inspiration for questions of your own devising to continue discussion with your students on the mathematics of this lesson.

Relation to Content Standards:

The goal of this eighth lesson is to think about irrational quantities and infinite polynomials (series). But it does have connections to some specific state standards.

High School: Proof and justification Classification of numbers Polynomial Theorems Finite and Infinite Series

Limits (convergence/divergence) (Grade 8 Existence of irrational numbers)

CONENTS:

Page 3:

High School PRACTICE PROBLEMS

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High School Grades

Irrational Numbers and Infinite Sums

Let’s start with infinite sums. As per usual, pick and choose which questions you’d like t try.

Question 1: In this picture of an 1 ← 𝑥𝑥 machine, what division problem is being solved and what is its answer?

Question 2: Use dots-and-boxes to show that 1

1 x+ equals 2 3 4 1 x x− + −x +x − . Question 3: Compute ₂ 1 x x

− . Do you get a sum of odd powers of x?

Question 4: We know 1 + 𝑥𝑥 + 𝑥𝑥2_{+ 𝑥𝑥}3_{… =} 1

1−𝑥𝑥 .

a) Use this to find a rational expression for 𝑥𝑥 + 𝑥𝑥2_{+ 𝑥𝑥}3_{+ 𝑥𝑥}4_….

b) Find a rational expression for _{1 a}₊ 3 ₊_a6 ₊_a9 ₊_a12 _+

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This next question is really cool! I advise you to draw very big boxes when you draw your dots and boxes picture. (The number of dots you need grows large quite quickly.)

Question 5: Compute 1 ₂

1 x x− − and discover the famous Fibonacci sequence! (What sequence of numbers is hidden in 1₂ ₃

1 x x− − −x ?)

Three optional questions for those who know calculus.

Question 6: Using _1−𝑥𝑥1 = 1 + 𝑥𝑥 + 𝑥𝑥2_{+ 𝑥𝑥}3_{…. , use Calculus to find an expansion for} 1

(1−𝑥𝑥)2.

Question 7: Use Calculus to find an expansion for _{(1−𝑥𝑥)}1 ₃. What special sequence are the coefficients?

Question 8: Using _1−𝑥𝑥1 = 1 + 𝑥𝑥 + 𝑥𝑥2_{+ 𝑥𝑥}3_{…. .}

a) Find a series for _1+𝑥𝑥1 ₂.

b) Use Calculus to find a series for arctan 𝑥𝑥.

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IRRATIONAL NUMBERS

Question 9: If one takes a 3-4-5 paper right triangle and folds its longest leg up to its

hypotenuse as shown, what would be the side lengths of the smaller copy of the right triangle that appears in top corner of the figure?

Question 10: A right triangle with integer side lengths has one leg k times as long as the other with k an integer. Call the side lengths band kb, and the hypotenuse a .

a) If we fold the longest leg up to the hypotenuse as shown, what are the side lengths of the smaller copy of the right triangle that appears in top corner of the figure? Are these lengths all integers?

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Question 11: If α is an irrational number, which of the following must be irrational too? a) 2α b) 1

α c) α d) 3α e) 2×α

Question 12: Euclid (300 B.C.) published 13 books called The Elements. In Volume III, he proved that √2 is irrational simply by reasoning about the evenness and oddness of numbers.

Can you fill in the details missing below?

Suppose that √2 is rational. What could be wrong with that?

We are assuming then that √2 =𝑎𝑎_𝑏𝑏 for positive integers a and b. In fact, we can assume this fraction is in its simplest form, meaning that and b have no common factor besides 1.

Squaring both sides give this the equation. ____________ Thus 2𝑏𝑏2 _{= 𝑎𝑎}2_.

So, 𝑎𝑎2_{is even. Now ask: Could a be odd? The answer is _____ because}

_____________________________________________________________. Hence, a is even and we can write 𝑎𝑎 = 2𝑐𝑐 for some whole number c. Squaring both sides of this equation gives ___________.

Substituting into our previous equation yields 2𝑏𝑏2 _{= 4𝑐𝑐}2_{. This simplifies to}

____________ telling us that 𝑏𝑏2_{is even. Consequently ______________ too.}

Hang on! This is a contradiction because ___________________________________.

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Question 13: Use a similar argument to Euclid to prove √5 is irrational. (But we can no longer use “evenness” in our proof. Can you figure out the details? What property about the number

5 do we have to rely on?)

Question 14: Suppose we wanted to prove √8 is irrational. How can you use what you know

about the irrationality of √2 to prove √8 is irrational?

Question 15: EXTRA

a) Look up the Rational Root Theorem. What does it say? b) Consider the polynomial p given by 𝑝𝑝(𝑥𝑥) = 𝑥𝑥2_{− 2.}

Show that the real number 𝑥𝑥 = √2 is a root of p.

c)_{Use the Rational Root Theorem to prove that √2 cannot be rational.}

Question 16: The decimal 0.30300300030000 … does not follow a repeating (since there is an

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BACKWARDS: IS EVERY REPEATING DECIMAL A FRACTION?

This is an interesting question.

Consider, for example, the repeating decimal 0.6363636.... Does this number arise from a division problem? That is, is this number a fraction?

A popular technique for attending to this issue starts by giving the quantity a name and to repeatedly multiply the quantity by ten. Let’s call the decimal C for Claude. (Why not?)

0. 63636363... 10 6. 36363636... 100 63. 6363636... C C C = = =

Let’s stop here since the infinite parts of C and 100C align perfectly. If we subtract them, we see

99C =63.6363663... 0.636363... 63− = and so 63 7

99 11

C = = , a fraction!

Question 17: a) Show that 0.213213213... is the fraction 213 999.

b) What fraction is 5.213213213... 5.213= ? c) What fraction is 2.8213213213... 2.8213= ?

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BONUS: CYCLIC NUMBERS

Question 19: The number N =142857 is called a cyclic number. (You might recognize that 1₇=. 142857��.)

a) Multiply 142857by each of 1, 2, 3, 4, 5, and 6. What do you notice? Why do you think this is called a cyclic number?

b) What is the decimal expansion of 10

7 ? Use this to explain why the decimal expansion of 3

7 must be 0. 428571�� .

c) Now, what is the decimal expansion of 30

7 ? Use this to explain why the decimal expansion of 2

7 must be 0. 285714�� . d) Now, what is the decimal expansion of 20

7 ? Use this quickly find the decimal expansion of 6

7.

e) Also, now quickly find the decimal expansions of 4 7 and

5 7.

Using the technique of the previous page and part b) of this question, we have 1 142857 7 999999= and 3 428571

7 999999= . This shows that 3N is indeed a “cycle” of the digits of N. Also,

2 285714 7 999999= shows that 2N is a cycle of the digits of N as well. And so on.

This explains why N , 2N, 3N, 4N, 5N, and 6N, are all cycles of the same digits.

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Question 20: Jezebel wonders if using the decimal expansion 1₈ yields another cyclic number for us. Does it?

Question 21: Zara suggests we use the decimal expansion of the number ₁₇1 to get another cyclic number. Use exploding dots to write ₁₇1 as an infinite repeating decimal. You should get a decimal that repeats every 16 digits! Check that this 16 digit number is indeed cyclic by

multiplying by 1, 2, 3,…, 16.

Mathematicians have proved that every cyclic number comes from a prime number p for which 1

p happens to have a decimal representation that cycles with period p − . (For 1

example, 7 is prime and 1

7has a decimal representation of period 6.) The work in the previous section and in question 19 shows that 1 ₁

999 99 10p 1

N N

p = _ = − ₋ , where N is cyclic

number p − digits long. 1

For the primes p = and 2 p = , the fractions 5 1 2 and

1

5 don’t have infinite decimal representations. For the prime p = , 3 1

3 has a periodic infinite decimal represent but with a period different from p − = . For 1 2 p = , 11 1

11 does not have period p − = . 1 10

Question 22: Check the decimal representation of 1

13. Does it have period 12?

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SOLUTIONS

1. After all the dot/antidot annihilations, we that we’re starting with a single dot. We’re looking for groups of x −1. It’s the problem

1

1 x− with answer

2 3

1 x x+ + +x _+. 2. Do it!

3. You get _{x x}₊ 3₊_x5₊_x7 _+_{, which makes}

sense as you can think of the expression as

( )

2 1 1 x x × − and this is

( ) ( ) ( )

(

₂ ₂ 2 ₂ 3

)

1 x× + x + x + x _+ .

4. a) We can see this is

(

₁ 2

)

₁ 1 ₁ 1 x x x + + + − = − −  or as

(

₁ 2

)

1 1 x x x x x + + + = × −  .

And can check that

1 ₁ 1 x− − is equivalent to 1 x x − . b) It’s

( ) ( )

3 3 2 3 1 1 1 a a a + + + = −  . c) It’s

(

1

)

1 1 1 x− − = x. 5. You get 2 2 3 4 5 6 1 1 1 2 3 5 8 13 x x x x x x x x − − = + + + + + + _+

This means, to study the properties of the Fibonacci sequence, you could study the properties of the rational function 1 ₂

1 x x− − instead!

Bonus Question: If you were to do all the

explosions in the infinite decimal

0.1|1| 2 | 3 | 5 | 8 |13 | 21| 34 | 55 | ... what number results? We have 2 3 2 3 4 5 6 1 1 1 2 4 7 13 24 x x x x x x x x x − − − = + + + + + + _+

This matches the sequence with every term past the first three is the sum of the three numbers before it.

6. Differentiate both sides to get

(

)

2 1 1 1 1 d dx x _x  _{ =}  ₋    − equal to 2 3 1 2+ x+3x +4x _+. 7. �_{(1−𝑥𝑥)}1 ₂�′ =_{(1−𝑥𝑥)}2 3 = 2 + 6𝑥𝑥 + 12𝑥𝑥2+ 20𝑥𝑥3_{+ ⋯.} Divide by 2 to see: 1 (1−𝑥𝑥)3= 1 + 3𝑥𝑥 + 6𝑥𝑥2+ 10𝑥𝑥3+ ⋯

The sequence 1, 3, 6, 10, … is the sequence of

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2 www.globalmathproject.org 8. a) _1+𝑥𝑥1 2= 1 − 𝑥𝑥2+ 𝑥𝑥4− 𝑥𝑥6+ 𝑥𝑥8+ ⋯ b) Integrate to see arctan 𝑥𝑥 + 𝐶𝐶 = 𝑥𝑥 −𝑥𝑥₃3+𝑥𝑥₅5−𝑥𝑥₇7+ ⋯ Putting x = 0 shows we must have C = 0. c) Put x = 1. Then

arctan 1 = 1 −1₃+1₅−1₇+ ⋯ So, 𝜋𝜋 = 4(1 −1₃+1₅−1₇+ ⋯ )

9. It’s a similar triangle with scale factor 1 3. It’s side lengths are 1, 4

3, and 5 3.

10. a) It’s a similar triangle and so has one leg k

times as long as the other.

The short leg is a kb− . Thus the second leg is

2

ka k b− . The hypotenuse is

(

2

) (

2 ₁

)

b ka k b− − = k + b ka− .

b) If k is an integer, all these lengths are integers too. So we have a smaller triangle with integer lengths. Repeating this indefinitely leads to an absurdity.

So, there can be no right triangle with integers sides b, kb, and a.That is, there are no integers satisfying _b2 ₊_{k b}2 2 ₌_a2_{. That is,}

there are no integers satisfying _k2 _{1 a}

b

+ = . No number of the form _{k +}2 ₁_{is rational.}

11. a) If 2 a b α = , then 2 a b α = is rational. Thus 2α cannot be rational. b) If 1 a b α = , then b a α = is rational. Thus 1 α cannot be rational. c) If a b α = , then a2₂ b α = is rational. Thus α cannot be rational. d) If 3 a b α = , then a3₃ b α = is rational. Thus 3_α _{cannot be rational.}

e) This can be rational. Have α = 2, for example.

12. 2 a2₂

b

= ; NO;

an odd number times itself is odd, not even;

2 ₄ 2

a = c ;

2 ₂ 2

b = c ;

bis even;

a is even and bis even, yet they are meant to only have 1 as a common factor.

13. If 5 a

b

= with a and bsharing no common factor (other than 1), then _a2 ₌₅_b2

shows that _a2_{is a multiple of}₅_{. Since}₅_is

prime, it seems this can only happen if a=5cis a multiple of 5. Then we get that ₂₅_c2 ₌₅_b2_,

that is, that _b2 ₌₅_c2_{, showing that}_b2_{, and}

hence b, is a multiple of 5 as well. Oops! That’s a contradiction.

BIG QUESTION: Is it obvious that if _n2_{is a}

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14. 8 2 2= . This is irrational by question 11. 15. a) The result states the possible rational roots p

q (assume this fraction is reduced) of a

polynomial 𝑎𝑎𝑥𝑥𝑛𝑛_{+ ⋯ + 𝑏𝑏 with integer} coefficients must be such that p is a factor (positive or negative) of b and q is a factor (positive or negative) of a.

b) 𝑝𝑝�√2� = �√2�2− 2 = 2 − 2 = 0 c) By Rational Root Theorem, if p

q is a root,

then p is a factor 2 (and so is either 1, 2, −1, or −2) and q is a factor of 1 (and so is either 1 or −1) . This means that the root p

q can only

possibly be 1, −1, 2, or −2. But none of these are actually roots! Since √2 is a root that means it cannot be a rational root. So 2is irrational. 16. Write down any infinitely long decimal expansion that you can be sure does not have a repeating pattern. Something like

0.121122111222111122221111122222... works. 17. a) Let A =.213213213.... Then 1000A =213.213213213... and 1000 213.213213213... 0.213213213... 213 A A− = − =

This gives 999A =213 and so 213 999 A = . b) 213 5208 5.213213213... 5 999 999 = + = . c) 8 .213213213... 2.8213213213... 2 10 10 8 213 2 10 9990 28185 9990 = + + = + + = 18. Set B =0.1| 1|1| 1|1| 1| ...− − − . Then 100B = −1| 1.1| 1|1| 1|1| 1| ...− − − and so 99B=100B B− = − =1| 1 10 1 9− = . Thus B = 9 1 99 11= .

(Society prefers us to unexplode each “1” and write 1 .0909090909....

11= .)

19. a) Each multiple has the same six digits, in order, but with a “wrap around” effect. b) We have 10 10 1 1.428571 7 = × =7 . But 10 1 3 1.428571 7 = + =7 showing that 3 0.428571 7 = . c) We have 30 10 3 4.285714 7 = × =7 and 30 ₄ 2

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4 www.globalmathproject.org d) 10 2 2.857142 7 × = and 20 2 6 7 = +7. Thus 6 0.857142 7 = . e) 60 8 4 8.571428 7 = + =7 gives 4 0.571428 7 = . 40 ₅ 5 _5.714285 7 = + =7 gives 5 0.714285 7 = . 20. It doesn’t. 1 0.125 8 = and there is on infinite cycle of repeating digits (except for repeating zeros) to help us out.

21. ₁₇1 = 0. 0588235294117647�� and 0588235294117647 is indeed cyclic.

22. 1