Is sqrt(6) an Irrational Number? A Proof without Prefix

  • Thread starter Mr. Fest
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In summary: Yes, thank you. I believe I mentioned I know that I can prove it in the same manner I can prove sqrt(2) is irrational. I just wanted to know if I can prove it using the fact that sqrt(6) = sqrt(2)*sqrt(3) and the fact that sqrt(2) is...well, you know.Ah, got it. Sorry for the confusion! In that case, yes, you can use the fact that sqrt(2) is irrational to prove that sqrt(6) is irrational. The key is to use proof by contradiction, as you pointed out earlier. So if you assume that sqrt(6) is rational, you can then use the fact that sqrt(
  • #1
Mr. Fest
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Homework Statement


Prove that [itex]\sqrt{6}[/itex] is irrational.

Homework Equations


The Attempt at a Solution


[itex]\sqrt{6}[/itex] = [itex]\sqrt{2}[/itex]*[itex]\sqrt{3}[/itex]

We know that [itex]\sqrt{2}[/itex] is an irrational number (common knowledge) and also this was shown in the textbook.

So, let's assume [itex]\sqrt{6}[/itex] and [itex]\sqrt{3}[/itex] are both rational.

[itex]\Rightarrow[/itex] [itex]\sqrt{2}[/itex]*[itex]\sqrt{3}[/itex] = [itex]\frac{a}{b}[/itex]

Also since, [itex]\sqrt{3}[/itex] is assumed to be rational it can be written as [itex]\frac{c}{d}[/itex] [itex]\Rightarrow[/itex] [itex]\sqrt{6}[/itex] = [itex]\sqrt{2}[/itex]*[itex]\frac{c}{d}[/itex] = [itex]\frac{a}{b}[/itex] [itex]\Leftrightarrow[/itex] [itex]\sqrt{2}[/itex] = [itex]\frac{d}{c}[/itex]*[itex]\frac{a}{b}[/itex] which is a contradiction since a, b, c and d are integers and [itex]\sqrt{2}[/itex] is not a rational number but [itex]\frac{d}{c}[/itex]*[itex]\frac{a}{b}[/itex] would be a rational number. Therefore, we can conclude that [itex]\sqrt{6}[/itex] is an irrational number.

Would this solution be correct? Or rather is this solution enough?
 
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  • #2
Not quite. Your proof has assumed something that isn't even true. [itex]\sqrt{3}[/itex] is irrational as you probably know, and you've only assumed it to be rational. If we consider that we didn't know whether it were rational or irrational, then we would need to include both cases in our proof.
 
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  • #3
Mentallic said:
Not quite. Your proof has assumed something that isn't even true. [itex]\sqrt{3}[/itex] is irrational and you probably know, and you've only assumed it to be rational. If we consider that we didn't know whether it were rational or irrational, then we would need to include both cases in our proof.

So that is needed? But if [itex]\sqrt{3}[/itex] is irrational then we have the product of two irrational numbers? Which per definition is irrational since the product of two different irrational numbers is irrational...?
 
  • #4
Mr. Fest said:
So that is needed? But if [itex]\sqrt{3}[/itex] is irrational then we have the product of two irrational numbers? Which per definition is irrational...?

The product of two irrational numbers can be rational: [itex]\sqrt{2} \times \sqrt{2} = 2[/itex].
 
  • #5
phyzguy said:
The product of two irrational numbers can be rational: [itex]\sqrt{2} \times \sqrt{2} = 2[/itex].

Yeah I know, that's why I wrote two different irrational numbers...
 
  • #6
Mr. Fest said:
Yeah I know, that's why I wrote two different irrational numbers...

[tex](\sqrt{2}+1)(\sqrt{2}-1)=1[/tex]
 
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  • #7
Mr. Fest said:
Yeah I know, that's why I wrote two different irrational numbers...

Exercise: Prove that given any rational number ##r\neq0## there are uncountably many pairs ##x,y## such that ##x## and ##y## are irrational, ##x\neq y##, and ##xy=r##.
 
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  • #8
gopher_p said:
Exercise: Prove that given any rational number ##r## there are uncountably many pairs ##x,y## such that ##x## and ##y## are irrational, ##x\neq y##, and ##xy=r##.

I wouldn't even know where to begin...

I'm in first year university...and there is a biiig different between high school and university over here. I basically aced every test in math during my high school years and I'm having trouble with this...
 
  • #9
Here's a hint. The proof that ##\sqrt{6}## is irrational has a proof that is very similar to the classic proof that ##\sqrt{2}## is irrational.
 
  • #10
Mr. Fest said:
I wouldn't even know where to begin...

I'm in first year university...and there is a biiig different between high school and university over here. I basically aced every test in math during my high school years and I'm having trouble with this...

Reframe the question: given a rational ##r\neq 0 ## and an irrational ##x##, is there a number ##y## such that ##xy=r##? Is this number rational or irrational?
 
  • #11
xiavatar said:
Here's a hint. The proof that ##\sqrt{6}## is irrational has a proof that is very similar to the classic proof that ##\sqrt{2}## is irrational.

I know how to prove ##\sqrt{2}## is irrational but I wanted to check if there was any way to just use the fact ##\sqrt{2}## is irrational and that ##\sqrt{6}## = ##\sqrt{2}##*##\sqrt{3}## to prove that ##\sqrt{6}## is irrational...
 
  • #12
gopher_p said:
Reframe the question: given a rational ##r\neq 0 ## and an irrational ##x##, is there a number ##y## such that ##xy=r##? Is this number rational or irrational?

No because if that was true then the irrational ##x## could be written as a rational, right?
 
  • #13
Mr. Fest said:
No because if that was true then the irrational ##x## could be written as a rational, right?

Theorem: if ##n > 0## is an integer and ##\sqrt{n}## is not an integer, then ##\sqrt{n}## is irrational. Google 'irrational square root'.
 
  • #14
Mr. Fest said:
No because if that was true then the irrational ##x## could be written as a rational, right?

Given an irrational number ##x## and a rational ##r\neq 0##, is ##rx^{-1}## a number? If it is a number, is it a rational number or an irrational number?

Just to be clear, this line of questioning has very little to do with the problem in the original post. It is directed towards the incorrect assumption that products of different irrationals must be irrational.
 
  • #15
Here is a start with proof by contradiction:
let's assume that the square root of 6 is rational.

By definition, that means there are two integers a and b with no common divisors where:

a/b = square root of 6.

So let's multiply both sides by themselves:

(a/b)(a/b) = (square root of 6)(square root of 6)
a2/b2 = 6
a2 = 6b2

But this last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least) is even. So a2 must be even. But any odd number times itself is odd, so if a2 is even, then a is even.

Can you go from here?
 
  • #16
Yellowflash said:
Here is a start with proof by contradiction:
let's assume that the square root of 6 is rational.

By definition, that means there are two integers a and b with no common divisors where:

a/b = square root of 6.

So let's multiply both sides by themselves:

(a/b)(a/b) = (square root of 6)(square root of 6)
a2/b2 = 6
a2 = 6b2

But this last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least) is even. So a2 must be even. But any odd number times itself is odd, so if a2 is even, then a is even.

Can you go from here?

Yes, thank you. I believe I mentioned I know that I can prove it in the same manner I can prove sqrt(2) is irrational. I just wanted to know if I can prove it using the fact that sqrt(6) = sqrt(2)*sqrt(3) and the fact that sqrt(2) is irrational...
 
  • #17
Can you try proving it by stating that rational numbers eventually have a periodic decimal expansion and irrationals do not?
 
  • #18
Irrationals are not closed under multiplication, so I do not think you can prove it like that.
 
  • #19
Yellowflash said:
Irrationals are not closed under multiplication, so I do not think you can prove it like that.

I already suggested to you in post #13 what you need to establish, and how to find it using Google. Have you even tried?
 
  • #20
Ray Vickson said:
I already suggested to you in post #13 what you need to establish, and how to find it using Google. Have you even tried?
Sorry Ray, totally forgot about that suggestion. Will look it up though! Thanks!
 

1. What does it mean for a number to be irrational?

An irrational number is a real number that cannot be expressed as a ratio of two integers. In other words, it cannot be written as a simple fraction. Irrational numbers are non-repeating and non-terminating decimals.

2. How do you prove that sqrt(6) is irrational?

To prove that sqrt(6) is irrational, we need to assume the opposite, that it is rational. This means that sqrt(6) can be expressed as a fraction a/b where both a and b are integers with no common factors. By squaring both sides, we get 6 = a^2/b^2. This means that a^2 is divisible by 6, and therefore a is also divisible by 6. This contradicts our assumption that a and b have no common factors, thus proving that sqrt(6) is irrational.

3. Can you use the same proof to show that sqrt(2) is irrational?

Yes, the same proof can be used to show that sqrt(2) is irrational. We assume that sqrt(2) is rational and follow the same steps to reach a contradiction.

4. What other methods can be used to prove the irrationality of a number?

Other methods include using the fundamental theorem of arithmetic, which states that every positive integer can be expressed as a unique product of prime numbers. If a number has a non-integer square root, it cannot be expressed as a product of integers and therefore is irrational. Another method is by using proof by contradiction, as shown in the proof for sqrt(6) above.

5. Why is proving the irrationality of a number important in mathematics?

Proving the irrationality of a number helps us understand the properties and relationships between different types of numbers. It also helps us develop new mathematical concepts and theories. Additionally, irrational numbers play a crucial role in fields such as geometry, physics, and engineering.

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