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

[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)[/color] 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!

 

[Dick]

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.

 

[kingwinner]

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
Contradiction!
So ¹⁷√(1/2) must be irrational. <====Is this proof correct?[/color]


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!

 

[NateTG]

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.

 

[Dick]

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
Contradiction!
So ¹⁷√(1/2) must be irrational. <====Is this proof correct?[/color]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?

 

[kingwinner]

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=p₁^a₁...p_k^a_k
m=q₁^b₁...q_k^b_k
where p_i, q_j are distinct primes
Then
p₁^17a₁...p_k^17a_k=2 q₁^17b₁...q_k^17b_k
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=p_i for some i. Also, 2 MAY occur in one of q_j
So left side we have 2^17a_i for some integer a_i
Right side we have 2 x 2^17b_j for some integer b_j
So it's impossible for these 2 numbers to be equal. Factorization into primes is unique, but 2^17a_i cannot be equal to 2 x 2^17b_j)
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!

 

[Dick]

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.

 

[kingwinner]

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

 

[NateTG]

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.

 

[kingwinner]

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[/color]. No prime factorizations exist, so the step in my proof
"...Now factor n and m into primes, say
n=p₁^a₁...p_k^a_k
m=q₁^b₁...q_k^b_k
where p_i, q_j are distinct primes"
is not valid. How can I actually prove that it also works in these 2 cases?

Can somebody help me, please?

 

[quantumdude]

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![/color]

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

Now, what is the fastest way to justify that 2√2, 2-√2, ¹⁷√(1/2)[/color] 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?

 

[kingwinner]

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![/color]

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]

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[/color]. No prime factorizations exist, so the step in my proof
"...Now factor n and m into primes, say
n=p₁^a₁...p_k^a_k
m=q₁^b₁...q_k^b_k
where p_i, q_j are distinct primes"
is not valid. How can I actually prove that it also works in these 2 cases?

Can somebody help me, please?

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

 

[rohanprabhu]

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?

 

[Gib Z]

Look here; https://www.physicsforums.com/showthread.php?t=195253

 

[Dick]

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?