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Quiz about The Square Root of the Problem

The Square Root of the Problem Quiz

Title comes from kyleisalive in the Author Challenges. This quiz walks the quiz taker through a proof that the square root of 3 is irrational. The proof is one that I wrote while taking a course in the history of mathematics.

A multiple-choice quiz by hlynna. Estimated time: 5 mins.

Author
hlynna

Time
5 mins

Type
Multiple Choice

Quiz #
333,825

Updated
Jul 23 22

# Qns
10

Difficulty
Average

Avg Score
7 / 10

Plays
555

One at a Time - Single Page - Timed Game

Question 1 of 10

1. This quiz uses sqrt as an abbreviation for square root and ^2 to indicate that a variable or number is squared. Our first step is to prove a lemma. What is a lemma? Hint

the proof of a statement needed to prove a theorem

bread made by Elves

a shaggy mammal with four legs

another word for theorem

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Question 2 of 10

2. Our first lemma:
Let x be an integer.
If x^2 is even, then x is even. I want to prove this by saying that if x is odd, then x^2 is odd. What method of proof is this? Hint

contradiction

conversion

contraposition

inversion

3. So, now I have to show that x is odd. I assume so, which means that for some integer we'll call n, x=2n+1. That means x^2=(2n+1)^2, I then manipulate the right half of that equation to get 4n^2+4n+1. What kind of expression is 4n^2+4n+1? Hint

A trinomial

A quadratic

A polynomial

All of these are correct

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Question 4 of 10

4. I factor a 2 out of the first two terms of 4n^2+4n+1, which gives me 2(2n^2+2n)+1. If I think of the part in parentheses as its own variable that I call m, then I have 2m+1. That must be an odd number! So, x is odd and I've proved my first lemma!

Whew. I'm not done yet, though. My proof requires one more lemma. Now I have to show that if x^2 is odd, then x is odd. I use contraposition again, saying that if x is even, then so is x^2. I assume x is even. Which of these is also true of x if it's even? Hint

x=2n, for any real number n

x=n-2, for some integer n

x=2n, for some integer n

x=n/2 for some integer n

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Question 5 of 10

5. Now I manipulate x=2n algebraically. First I square both sides. Then I further manipulate (2n)^2. This simplifies to 4n^2, and I can factor out a 2 to get 2(2n^2). If I once again see the part in parentheses as its own thing, I can call it m. Now I have x^2=2m. Therefore, x^2 is even. This proves my second lemma!

Now I can really move into the proof that sqrt3 is irrational. To begin with, what is an irrational number? Hint

A number that makes no sense

A number that does not appear on the number line

A number that does not exist

A number that cannot be written as the ratio of two integers

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Question 6 of 10

6. Since I want to show that sqrt3 is irrational, a good start would be to assume that it's rational and see if there's a contradiction. If a number is rational, then it can be written as the ratio of two integers, so I say that sqrt3=a/b, where a and b are integers. What number can b never be? Hint

one

a lonely number

any negative number

zero

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Question 7 of 10

7. I am also going to assume that my ratio of a/b is also in its lowest terms. I lose no generality here. My proof says, "WLOG we may assume gcd(a,b)=1."

If the d in gcd stands for "divisor," what does gcd stand for? (Hint: You've probably seen the same kind of abbreviation with an F which stood for factor.) Hint

greatest correlative divisor

general correlative divisor

greatest common divisor

general common divisor

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Question 8 of 10

8. So, I have sqrt3=a/b. I'll do some algebra! I multiply both sides by b, then square both sides. Now I have 3b^2=a^2.

Now I have two cases to consider. One is that a^2 is even. What is the other? Hint

that a^2 is odd

that 3b^2 is odd

that 3b^2 is even

that 3b^2 does not equal a^2

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Question 9 of 10

9. In my first case, I consider a^2 as even. My first lemma says that means a is even. Also, I know that 3b^2 is even. Therefore, b^2 is even, which means that by Lemma 1, b is even.

Therefore, gcd(a,b) does not equal 1. I have a contradiction!

What is the common divisor that causes that contradiction? Hint

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Question 10 of 10

10. Well, one more case to contradict so I can show that sqrt3 is irrational! Here I consider that a^2 could be odd. My second lemma says then a is odd. That means that a=2c+1, where c is an integer.

I also know that 3b^2 is odd, which means b^2 is odd and that b is odd. So, I'll say that b=2d+1, for some integer d.

Now this is more complicated algebra. Since a^2=3b^2, then (2c+1)^2=3(2d+1)^2. Several lines of algebraic manipulation yield this: 2(c^2+c)=2(3d^2+3d)+1. The left half is always even and the right half is always odd. They definitely don't equal each other! I have another contradiction!

Both cases result in a contradiction. So, sqrt3 must not be rational. Therefore, sqrt3 is irrational. What symbol could I use to say "therefore?" Hint

a fancy th

three dots in a triangle

QED

a fancy TF

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Quiz Answer Key and Fun Facts

1. This quiz uses sqrt as an abbreviation for square root and ^2 to indicate that a variable or number is squared. Our first step is to prove a lemma. What is a lemma?

Answer: the proof of a statement needed to prove a theorem

Sometimes before we can prove a theorem, we must first prove a statement that the proof of our theorem depends on. These statements are called lemmas.

The shaggy mammal is a llama and the bread made by Elves is lembas (from the "Lord of the Rings").

2. Our first lemma: Let x be an integer. If x^2 is even, then x is even. I want to prove this by saying that if x is odd, then x^2 is odd. What method of proof is this?

Answer: contraposition

The contrapositive of a statement is true if the original statement is true. If I say that all cows eat grass (thus evoking a mnemonic from music), I can rewrite that as "If an animal is a cow, then it eats grass." The contrapositive of this statement is "If that animal is not eating grass, then it is not a cow."

The other three are also ways of changing a statement in logic.

(Thanks to wikipedia for reminding me of just how contraposition works!)

Answer: All of these are correct

A polynomial is any expression which consists of variables and constants where the operations are addition, subtraction, multiplication, and raising to a positive power. We can refer to an expression with three terms as a trinomial; 4n^2+4n+1 has three terms. Furthermore, if the highest power in a polynomial is 2, then it is quadratic.

4. I factor a 2 out of the first two terms of 4n^2+4n+1, which gives me 2(2n^2+2n)+1. If I think of the part in parentheses as its own variable that I call m, then I have 2m+1. That must be an odd number! So, x is odd and I've proved my first lemma! Whew. I'm not done yet, though. My proof requires one more lemma. Now I have to show that if x^2 is odd, then x is odd. I use contraposition again, saying that if x is even, then so is x^2. I assume x is even. Which of these is also true of x if it's even?

Answer: x=2n, for some integer n

In layman's terms, any even number is equal to twice some number. However, in math, we must be careful to say what kind of number. Here, we are dealing with integers, i.e., the positive and negative numbers and 0. If we say all real numbers, then we get fractions and numbers like pi, which we can certainly multiply by two, but they won't necessarily be odd or even.

The other options where n is an integer will not always yield an even number for x. In x=n-2, if n=3, then x is odd, just as an immediate counterexample. In x=n/2, if n=3, then x is 3/2, which is neither odd nor even.

5. Now I manipulate x=2n algebraically. First I square both sides. Then I further manipulate (2n)^2. This simplifies to 4n^2, and I can factor out a 2 to get 2(2n^2). If I once again see the part in parentheses as its own thing, I can call it m. Now I have x^2=2m. Therefore, x^2 is even. This proves my second lemma! Now I can really move into the proof that sqrt3 is irrational. To begin with, what is an irrational number?

Answer: A number that cannot be written as the ratio of two integers

There are many irrational numbers. Pi and e are the most widely known and a great many square roots (and higher roots) are irrational, as well.

Irrational numbers do appear on a number line. For example, pi is a bit more than 3 and would have a tick mark on a number line just to the right of 3. The number e would be past 2, and close to 3.

Answer: zero

We can never divide by zero (I like to say that that's where black holes come from, although I know that's not true. It does amuse my students, though.).

We show that we remember this important rule in proofs by noting that b does not equal 0 in the same line we would say a and b are integers.

7. I am also going to assume that my ratio of a/b is also in its lowest terms. I lose no generality here. My proof says, "WLOG we may assume gcd(a,b)=1." If the d in gcd stands for "divisor," what does gcd stand for? (Hint: You've probably seen the same kind of abbreviation with an F which stood for factor.)

Answer: greatest common divisor

Think about it;, if a fraction's numerator and denominator have a divisor in common, it's not in lowest terms. The fraction 2/4 can be divided by 2 to get 1/2. One and two share only one divisor, 1, so 1/2 is in lowest terms.

"WLOG" stands for "without loss of generality." Generality is very important in proofs.

8. So, I have sqrt3=a/b. I'll do some algebra! I multiply both sides by b, then square both sides. Now I have 3b^2=a^2. Now I have two cases to consider. One is that a^2 is even. What is the other?

Answer: that a^2 is odd

Conveniently, I have two lemmas that deal with my two cases!

9. In my first case, I consider a^2 as even. My first lemma says that means a is even. Also, I know that 3b^2 is even. Therefore, b^2 is even, which means that by Lemma 1, b is even. Therefore, gcd(a,b) does not equal 1. I have a contradiction! What is the common divisor that causes that contradiction?

Answer: 2

If you have a ratio where both terms are even, they have a common divisor of 2. All even numbers are divisible by 2. Since I created a ratio a/b where I could show that both a and b were even, I knew that ratio wasn't in lowest terms. In my earlier statements, I said that gcd(a,b)=1. I contradicted that, which means case #1 will help me in my proof. Case #1 basically half-proves that sqrt3 isn't rational, which means it must be irrational.

10. Well, one more case to contradict so I can show that sqrt3 is irrational! Here I consider that a^2 could be odd. My second lemma says then a is odd. That means that a=2c+1, where c is an integer. I also know that 3b^2 is odd, which means b^2 is odd and that b is odd. So, I'll say that b=2d+1, for some integer d. Now this is more complicated algebra. Since a^2=3b^2, then (2c+1)^2=3(2d+1)^2. Several lines of algebraic manipulation yield this: 2(c^2+c)=2(3d^2+3d)+1. The left half is always even and the right half is always odd. They definitely don't equal each other! I have another contradiction! Both cases result in a contradiction. So, sqrt3 must not be rational. Therefore, sqrt3 is irrational. What symbol could I use to say "therefore?"

Answer: three dots in a triangle

Here is the full algebraic manipulation, left out of the question so your eyes don't bleed.

4c^2+4c+1=3(4d^2+4d+1)
4c^2+4c+1=12d^2+12d+3
2(2c^2+2c)+1=2(6d^2+6d)+3
2(2c^2+2c)=2(6d^2+6d)+2
2c^2+2c=6d^2+6d+1
2(c^2+c)=2(3d^2+3d)+1

Source: Author hlynna

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