# Proving Irrationality of 2√2, 2-√2, 17√(1/2)

• kingwinner
In summary, the proofs for irrationality of 2√2, 2-√2, and 17√(1/2) can be justified by constructing polynomial equations and applying the Rational Root Theorem (RRT). This can be done due to the fact that the square root of 2 is irrational, and therefore, any expressions involving it will also be irrational.
kingwinner
I know that √2 is irrational (and I've seen the proof).

Now, what is the fastest way to justify that 2√2, 2-√2, 17√(1/2) are irrational? (they definitely "seem" to be irrational numbers to me) Can all/any these follow immediately from the fact that √2 is irrational?

Thanks!

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The product of an irrational and a nonzero rational is always irrational. Prove this, it's not hard. The sum of an irrational and a rational is always irrational. Prove this, also not hard. This last one is not completely trivial. You just have to think about prime factorization. It's basically the same proof as that the sqrt(2) is irrational.

Dick said:
The product of an irrational and a nonzero rational is always irrational. Prove this, it's not hard. The sum of an irrational and a rational is always irrational. Prove this, also not hard. This last one is not completely trivial. You just have to think about prime factorization. It's basically the same proof as that the sqrt(2) is irrational.
Let me try the last one,
Suppose
17√(1/2)=m/n
=> n17=2 m17
By unique prime facotrization, 2 must occur as a prime factor of n (since the right side has a "2"), so on left side, 2 to the exponent a multiple of 17
On the right side, 2 to the exponent (a multiple of 17) + 1
So 17√(1/2) must be irrational. <====Is this proof correct?

On the other hand, I am still having trouble with understanding why 2√2 and 2-√2 are irrational. How can you know for sure that
"The product of an irrational and a nonzero rational is always irrational"
& "The sum of an irrational and a rational is always irrational" ?
I have no idea how to prove those...

Thanks for explaining!

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Assume, by contradiction that $2\sqrt{2}$ is rational, then you can write it as:
$\frac{a}{b}$
with integers $a,b$, but then
$\sqrt{2}=\frac{a}{2b}$
contradicts that $\sqrt{2}$ is irrational.

kingwinner said:
Let me try the last one,
Suppose
17√(1/2)=m/n
=> n17=2 m17
By unique prime facotrization, 2 must occur as a prime factor of n (since the right side has a "2"), so on left side, 2 to the exponent a multiple of 17
On the right side, 2 to the exponent (a multiple of 17) + 1
So 17√(1/2) must be irrational. <====Is this proof correct?On the other hand, I am still having trouble with understanding why 2√2 and 2-√2 are irrational. How can you know for sure that
"The product of an irrational and a nonzero rational is always irrational"
& "The sum of an irrational and a rational is always irrational" ?
I have no idea how to prove those...

Thanks for explaining!

You aren't quite getting the 2^(1/17) proof either. Look up and understand the sqrt(2) proof. You HAVE to say "assume 2^(17)=m/n is in lowest terms", i.e. gcd(m,n)=1. Now you say 2*n^17=m^17. So m is even. So 2^17 divides 2*n^17. So 2 divides n. So n is even. Now what?

Dick said:
You aren't quite getting the 2^(1/17) proof either. Look up and understand the sqrt(2) proof. You HAVE to say "assume 2^(17)=m/n is in lowest terms", i.e. gcd(m,n)=1. Now you say 2*n^17=m^17. So m is even. So 2^17 divides 2*n^17. So 2 divides n. So n is even. Now what?
But I don't see where is it going wrong in the logic of my proof, would you (or anybody else) mind pointing it out, please?

Proof:
Suppose 17√(1/2)=m/n
=> n17 = 2 m17
Now factor n and m into primes, say
n=p1a1...pkak
m=q1b1...qkbk
where pi, qj are distinct primes
Then
p117a1...pk17ak=2 q117b1...qk17bk
If these are equal, then since 2 occurs in the prime facotrization on the right side, it must occur on the left side (by uniqueness of prime factorization), so 2=pi for some i. Also, 2 MAY occur in one of qj
So left side we have 217ai for some integer ai
Right side we have 2 x 217bj for some integer bj
So it's impossible for these 2 numbers to be equal. Factorization into primes is unique, but 217ai cannot be equal to 2 x 217bj)
So 17√(1/2) must be irrational.

And my method also seems to work in a more general situation like proving that 3√(2/7) is irrational, too. But the method of proving that √2 is irrational doesn't seem to work here...

Please let me know if I am wrong anywhere.

Thanks!

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I believe you. 17*ai=17*bi+1 doesn't work. 1 isn't divisible by 17. But it really is the same proof as for sqrt(2). In that case you get 2*ai=2*bi+1. Still doesn't work. 1 isn't divisible by 2 either.

Is there a neat way to prove that 3√(2/7) is irrational?

kingwinner said:
Is there a neat way to prove that 3√(2/7) is irrational?

You might as well just move on to a generic proof for all roots of rational numbers.

In wikipedia, there is a method using the concept of unique prime facotraztion to prove that a number is irrational, so I think my proof on 17√(1/2) being irrational is also correct.

http://en.wikipedia.org/wiki/Square_root_of_two#Proof_by_unique_factorization

17√(1/2)=m/n
But I noticed that there are two BAD cases: m=1 or n=1. No prime factorizations exist, so the step in my proof
"...Now factor n and m into primes, say
n=p1a1...pkak
m=q1b1...qkbk
where pi, qj are distinct primes"
is not valid. How can I actually prove that it also works in these 2 cases?

kingwinner said:
I know that √2 is irrational (and I've seen the proof).

The proof? There isn't just one!

Are you aware of the following result?

Theorem: The Rational Root Theorem (RRT)
Let $a_nx^n+...+a_1x+a_0=0$ be a polynomial equation with $a_i\in\mathbb{Z}$ for all $i$, and with $a_n,a_0\neq 0$. Then $p/q$ be a rational root for the equation with $p$ relatively prime to $q$ and $q\neq 0$.

Then:

(i) $p$ divides $a_0$, and
(ii) $q$ divides $a_n$.

Now consider the equation $x^2-2=0$, which clearly has $\sqrt{2}$ as a root. By RRT, the only possible rational roots are -2,-1,1,2. But none of them works!

Therefore, $\sqrt{2}$ is irrational.

Now, what is the fastest way to justify that 2√2, 2-√2, 17√(1/2) are irrational? (they definitely "seem" to be irrational numbers to me) Can all/any these follow immediately from the fact that √2 is irrational?

I don't know about the fastest way, but a viable way would be to construct polynomial equations as I have done above, and apply RRT. Here, I'll do another one for you.

Let $x=2\sqrt{2}$. Then $x^2=8$, or $x^2-8=0$.

Clearly, this equation has $2\sqrt{2}$ as a root. By RRT, the only possible rational roots to that equation are -8,-4,-2,-1,1,2,4,8. But none of these works. Therefore, $2\sqrt{2}$ is irrational.

Comprede?

Tom Mattson said:
The proof? There isn't just one!

Are you aware of the following result?

Theorem: The Rational Root Theorem (RRT)
Let $a_nx^n+...+a_1x+a_0=0$ be a polynomial equation with $a_i\in\mathbb{Z}$ for all $i$, and with $a_n,a_0\neq 0$. Then $p/q$ be a rational root for the equation with $p$ relatively prime to $q$ and $q\neq 0$.

Then:

(i) $p$ divides $a_0$, and
(ii) $q$ divides $a_n$.

Now consider the equation $x^2-2=0$, which clearly has $\sqrt{2}$ as a root. By RRT, the only possible rational roots are -2,-1,1,2. But none of them works!

Therefore, $\sqrt{2}$ is irrational.

I don't know about the fastest way, but a viable way would be to construct polynomial equations as I have done above, and apply RRT. Here, I'll do another one for you.

Let $x=2\sqrt{2}$. Then $x^2=8$, or $x^2-8=0$.

Clearly, this equation has $2\sqrt{2}$ as a root. By RRT, the only possible rational roots to that equation are -8,-4,-2,-1,1,2,4,8. But none of these works. Therefore, $2\sqrt{2}$ is irrational.

Comprede?

Sorry about that, but I thought that was only one way before, since only one proof in presented in my class...but now I am aware of many proofs...

kingwinner said:
In wikipedia, there is a method using the concept of unique prime facotraztion to prove that a number is irrational, so I think my proof on 17√(1/2) being irrational is also correct.

http://en.wikipedia.org/wiki/Square_root_of_two#Proof_by_unique_factorization

17√(1/2)=m/n
But I noticed that there are two BAD cases: m=1 or n=1. No prime factorizations exist, so the step in my proof
"...Now factor n and m into primes, say
n=p1a1...pkak
m=q1b1...qkbk
where pi, qj are distinct primes"
is not valid. How can I actually prove that it also works in these 2 cases?

Does anyone have any idea how to deal with the case m=1 or n=1? I am stuck here...

kingwinner said:
I know that √2 is irrational (and I've seen the proof).

sorry for being offtopic and interfering.. but how can one proove that √2 is irrational?

kingwinner said:
Does anyone have any idea how to deal with the case m=1 or n=1? I am stuck here...

(1/2)^(1/17)=1/n -> 2=n^(17). That's ridiculous. 2 isn't the 17th power of any integer. (1/2)^(1/17)=m/1 -> 2*m^(17)=1. Also ridiculous. 1/2 isn't the 17th power of any integer. Was that really so mysterious?

## What does it mean for a number to be irrational?

A number is considered irrational if it cannot be written as a ratio of two integers (a fraction) and its decimal expansion is non-terminating and non-repeating.

## How can the irrationality of a number be proven?

One way to prove the irrationality of a number is to show that it cannot be expressed as a fraction and its decimal expansion does not have a pattern or repetition.

## What is the significance of proving the irrationality of 2√2, 2-√2, 17√(1/2)?

Proving the irrationality of these numbers helps to strengthen the understanding of irrational numbers and their properties. It also reinforces the fact that not all numbers can be represented as fractions.

## What is the proof for the irrationality of 2√2, 2-√2, and 17√(1/2)?

The proof for the irrationality of these numbers involves assuming that they can be expressed as fractions and then using algebraic manipulations to arrive at a contradiction.

## Are there any other methods for proving the irrationality of numbers?

Yes, there are other methods such as using the irrationality of √2 and √3 to prove the irrationality of other numbers, or using the fact that the square root of a prime number is irrational. There are also more advanced mathematical methods, such as using continued fractions, to prove the irrationality of numbers.

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