Is sqrt(5) Rational? Understanding the Proof by Contradiction

[Klion]

Alright, heading says it all. This is a nice problem heh.. I can see how to prove sqrt(5) is irrational. I think this method works up to the points where the fact 5 is a prime is used, (ie prime lemma) on 5 which doesn't work so well on 6! hehe

Was thinking of maybe using product of primes somehow but.. hmm dunno

Anyway for sqrt(5) went like this (proof by contradiction)

Prove sqrt(5) is not rational.

Suppose sqrt(5) = a/b, where a,b E Z+ (suppose sqrt(5) is rational)
we'll also assume gcd(a,b) = 1 (otherwise just divide a,b by gcd)
sqrt(5) = a/b <=> 5 = a^2/b^2 or 5*b^2=a^2
so 5|a^2, but 5 is a prime so 5|a Ea' 5*a'=a so 5b^2=(5a)^2 or 5b^2=5^2*a^2 or b^2 = 5*a'^2 so 5|b^2 by prime lemma 5|b 5|a ^ b|b so 5|gcd(a,b)
so 5|1 which is nonsense therefore our initial assumption is incorrect and 5 is in fact irrational.

-Kli

 

[KLscilevothma]

Suppose sqrt(6) is rational

gcd(a,b) = 1

\sqrt{6} = \frac{a}{b}

6b^2 = a^2

Now there are 4 possibilities.

1) both a and b are even.
It cannot be true since gcd(a,b)=1

2) a is odd and b is even
It cannot be true too. If a is odd, then left hand side will be even while right hand side will be odd.

3) both a and b are odd.
It cannot be true too and the reason is the same as that in (2)

4) a is even and b is odd.

Let a = 2m and b = 2n + 1

6b^2 = a^2

6*{(2n+1)}^2 = {2m}^2

12m^2 + 12m +3 = 2n^2

left hand side is odd while right hand side is even

Contradiction.

Therefore sqrt(6) is irrational

 

[HallsofIvy]

"Let a = 2m and b =2n + 1
6b²= a²

6(2n+1)²= 2m²"

No, it should be 6(2n+1)²= (2m)²

"12m²+ 12m+ 3= 2n²"

Now this completely baffles me. Even assuming you accidently switched m and n, it doesn't follow.
From 6(2n+1)²= (2m)² you get
6(4n²+ 4n+ 1)= 4m² or
24n²+ 24n+ 6= 4m² and both sides are clearly even.

 

[KLscilevothma]

Originally posted by HallsofIvy
24n²+ 24n+ 6= 4m² and both sides are clearly even.

How about dividing both sides by 2, then we get
12n²+ 12n+ 3= 2m²
where left hand side is odd while right hand side is even.

If we don't divide both sides by 2, left hand side cannot be divisible by 4 while right hand side can.

 

[mathman]

How about the following (for any n not a square of an integer).

sqrt(n)=a/b, where a and b are integers with no common factors, and b >1.

n=a²/b²

The numerator and denominator still have no common factors, so n is not an integer.

 

[phoenixthoth]

...for any n not a square of an integer...n is not an integer.

any n in what? is n any nonnegative rational number, then? if it's any nonnegative integer n not the square of an integer, it seems odd to end by saying "n is not an integer."

 

[mathman]

Sorry for the confusion. I assumed we were discussing integers in general. My discussion is about (positive) integers which are not squares of integers. In n is a square of an integer, then b=1.

 

[phoenixthoth]

then wouldn't "n is not an integer" prove it?

 

[uart]

Originally posted by phoenixthoth
then wouldn't "n is not an integer" prove it?

Hi phoenixthoth, Mathman is using proof by contradiction . In the first line of his proof what he means is : Assume that n is a positive integer that is not a perfect square.

He goes on to show that assuming sqrt(n) to be rational, but non integer (see b>1), leads directly to a contradiction of the assumption that n was an integer. Hence the result that the square root of all integers, other than the perfects squares, are irrational. It's a perfectly valid proof.

 

[phoenixthoth]

it seems fishy because it was easier to prove it in general for nonsquare nonnegative integers n than for n=6 and it's usually the case that it's easier to do it for n=6 and harder in general. I'm at some point if i feel like it going to check where the assumption that n is not a square was used and if this proof applies to square integers as well. i suppose that's what enables one to say b>1.

 

[phoenixthoth]

and 4 = 2*2 both of which are prime

 

[selfAdjoint]

What has 4 = 2 X 2 got to do with that? Some squares are squares of primes and some aren't, like 6 X 6 = 36. How does that affect the proof?

 

[kx21]

...

(Repost of 'disappeared' Post with Refinement)

Proof;-

1) Each SQRT(Prime Number) is irrational &

2) SQRT(Prime Number 1) * SQRT(Prime Number 2) is irrational if Prime Number 1 is not equal to Prime Number 2.

3) SQRT(6) = SQRT(2) * SQRT(3)

4) 2 & 3 are two different Prime Numbers.

i.e.

5) SQRT(6) is irrational.

This completes the proof...


kx21

https://www.physicsforums.com/member.php?s=&action=getinfo&find=lastposter&forumid=80

 

[phoenixthoth]

Originally posted by selfAdjoint
What has 4 = 2 X 2 got to do with that? Some squares are squares of primes and some aren't, like 6 X 6 = 36. How does that affect the proof?

this was in reference to a deleted post.

 

[phoenixthoth]

Originally posted by kx21
...

(Repost of 'disappeared' Post with Refinement)

Proof;-

1) Each SQRT(Prime Number) is irrational &

2) SQRT(Prime Number 1) * SQRT(Prime Number 2) is irrational if Prime Number 1 is not equal to Prime Number 2.

3) SQRT(6) = SQRT(2) * SQRT(3)

4) 2 & 3 are two different Prime Numbers.

i.e.

5) SQRT(6) is irrational.

This completes the proof...


kx21

https://www.physicsforums.com/member.php?action=getinfo&find=lastposter&forumid=111&

assumption #1 is as clear as conclusion #5. some people would not take #1 without proof.

selfadjoint, this post in its original form was about six words long and the point in bringing up 4=2x2 is that it should have been stated that the two primes must be different; it was stated in the second form. this argument would have to be messaged before it would work for all n. the main thing there would be to prove/say a nonsquare has not all even powers in its prime factorization.

 

[uart]

1) Each SQRT(Prime Number) is irrational &

2) SQRT(Prime Number 1) * SQRT(Prime Number 2) is irrational if Prime Number 1 is not equal to Prime Number 2.

The problem is that you never really establish the truth of either point 1 or point 2 anywhere in the proof kx21


I still think that the best proof is as follows.

Let \sqrt{n} = \frac{a}{b}

Then a^2 = b^2 n <- Eqn1

From the fundamental therom of arithmetic (that is the uniqueness of prime factorization) it is clear that any integer squared contains only even powered prime factors.

Thus the LHS of Eqn1 can contain only even powered prime factors whereas the RHS of Eqn1 will contain odd powered prime factors whenever n contains any odd powered prime factors.

So LHS = RHS is possible if (and only if) n contains only even powered prime factors. This is of course equivalent to the requirement that n be a perfect square.

In other words, either n is a perfect square or \sqrt{n} is irrational.

 

[kx21]

*** a 'M' Generalization with a bonus Link Phi, e & i... ***

Theorem- The irrationality of SQRT(N)

SQRT(N) is irrational if

N = p1*p2...*pk, where

pi i =1 to k are different Prime Numbers.


For instances,

1) Is SQRT(6) irrational?

...

It's irrational as 6 = 2 * 3

[?]


2) Is SQRT(30) irrational?

A simple Number Test,

with a Bonus Link about Phi (Golden Ratio), e, pi for all:-

http://superstringtheory.com/forum/extraboard/messages12/376.html

Happy exploring & have a wonderful memory & time...

kx21

https://www.physicsforums.com/member.php?s=&action=getinfo&find=lastposter&forumid=80

 

[uart]

Theorem- The irrationality of SQRT(N)

SQRT(N) is irrational if

N = p1*p2...*pk, where

pi i =1 to k are different Prime Numbers.

While correct that theorem is only "half baked" as there are very many integers n for which sqrt(n) is irrational but yet n cannot be expressed as such a simple prime product as stated by the therom (that is, with all prime factors to unity power only).

Why not do it properly as per either mathmans or my own post, and establish irrationality for sqrt of all integers which are not perfect squares ?


Also, if you're going to just quote a therom that has already established primeness of certain square roots (as opposed to establishing it from something more fundamental like the uniqueness of prime factorization) then why pick such a "half baked" theorem in the first place? Why not use a better theorem to begin with.


Theorem : Either n is a perfect square or sqrt(n) is irrational.

Then all you need to do is to say. "Since 6 in not a perfect then by the above theorem then the sqrt of 6 is irrational." QED.

Really, if you are going to be lame then at least do it properly. ;-)

 

[kx21]

Originally posted by uart

Theorem : Either n is a perfect square or sqrt(n) is irrational.

Then all you need to do is to say. "Since 6 in not a perfect then by the above theorem then the sqrt of 6 is irrational." QED.

Really, if you are going to be lame then at least do it properly. ;-)

...

Let me think...

Are there any 'Topological Defects' hidden in your Theorem.

Eureka!

Your Theorem is invalid if...

n = -1.

Proof:-

Given SQRT(-1)= i * i

This Completes the proof...


[?]


kx21
https://www.physicsforums.com/member.php?s=&action=getinfo&find=lastposter&forumid=80

 

[phoenixthoth]

it was clarified earlier that n is a nonnegative nonsquare integer. i is irrational if irrational means not rational. but irrational usually refers to R\Q. technically, this doesn't violate an or statement though i think what was mean by adding the word "either" was exclusive or, Xor.

 

[oen_maclaude]

square(6) is irrational.

first of all, we let x = sqrt(6).
squaring both sides we have xsquare is equal to 6.

 

[HallsofIvy]

Okay, but that requires proving the "rational root theorem" which is surely harder to prove in general that specifically that sqrt(6) is irrational.

 

[oen_maclaude]

well, i am sorry but i haven't heard of the prime lemma. however, in my convinience, it is better to prove the statement using the rational root theorem. the only thing that you need here is the rational root theorem.

 

[modmans2ndcoming]

proove that the SR of 2 is Irrational, and the SR or 3 is irrational (they are prime so they have no factors other than 1 and itself therefore the root is irrational)

since those two roots are irational, an Irrational times an irrational is an irrational, therefore, root 6 is irrational.

 

[modmans2ndcoming]

your proof using the rational root theorim is incomplete.

you need to point out that since 6 is a real, Positive number, the SQRT of 6 cannot be complex, so the roots of X^2 can not be complex. (the complex root theorim says complex roots come in pairs, so it is important to rule those root out)

 

[matt grime]

Originally posted by modmans2ndcoming
proove that the SR of 2 is Irrational, and the SR or 3 is irrational (they are prime so they have no factors other than 1 and itself therefore the root is irrational)

since those two roots are irational, an Irrational times an irrational is an irrational, therefore, root 6 is irrational.

certainly not true that last bit. It is easily possible for the product of two irrationals to be rational, sqrt(2) and, er, sqrt(2) for instance.

 

[modmans2ndcoming]

sorry, I was not percise enough in my proof and left some speculation.

I should have said " the product of 2 distinct irrational numbers is irrational.

 

[matt grime]

Originally posted by modmans2ndcoming
sorry, I was not percise enough in my proof and left some speculation.

I should have said " the product of 2 distinct irrational numbers is irrational.

and the product of pi and 1/pi is?

 

[modmans2ndcoming]

umm...1/pi is not an integer so it can't be a prime :-)

 

[matt grime]

you didn't state that it *had* to be only to do with integers or primes. You just said that the product of two (distinct) irrationals is irrational.

If you don't like pi, then sqrt(2) and 2sqrt(2) are two distinct irrationals involving only integers and whose product is rational. Until such time as you accurately state all your hypotheses and what that impies you can't discount counter examples. So phrase the statement to remove ambiguity.

 

[modmans2ndcoming]

well first off, the discussion was in the universe of Natural numbers since we are discussing primes.

so here it is.

6 is a natural number. six can be composed as 2*3. 2 and 3 are primes, so the SQRT of a prime is irrational since a prime has no factors other than itself and 1. if you multiply the root of 2 distict prime numbers together, you end up with an irrational number, and since SQRT(6) can be composed as SQRT(2*3) and since we know that the SQRT of 2 and 3 are irrational and that the product of the SQRT of 2 distict prime numbers is irrational, SQRT(6) is irrational.

there, I have used all the language in one context.

 

[matt grime]

Thats fine. At no point did you state that your deductions were dependent on being only taking roots of primes though. And the result you gave had been proven on page 1 of the thread by two or three people without these imprecisions.

 

[NateTG]

if you multiply the root of 2 distict prime numbers together, you end up with an irrational number

Of course, this is probably what the original poster was trying to prove.

If you want to do something like this, it's easier and more understandable to do this:

Assume the fundamental theorem of algebra:
Each natural number has a unique prime factorization.

And then prove the theorem:
Any number whose prime factorization has any odd exponents has an irrational square root.

Proof:
Assume by contradiction that
\sqrt{n}=\frac{a}{b}
where \frac{a}{b} is a reduced fraction.
We can square both sides of the equation to get:
n=\frac{a^2}{b^2}
Now, we know that
n=mp^{2k+1}
where m is not divisible by p for some prime p by the hypothesis.
So we have
mp^{2k+1}=\frac{a^2}{b^2}
so
b^2mp^{2k+1}=a^2
Now we have 2 cases:
case 1:
p divides a
Then b is not divisible by p since the fraction is in reduced form.
Then a=p^i a&#039; where a&#039; is not divisble by p.
so
b^2mp^{2k+1}=(p^ia&#039;)^2
b^2mp^{2k+1}=p^{2i}{a&#039;}^2
Now the prime factorization of the LHS will have an odd power of p (2k+1) and the RHS will have an even power of p (2i). So by the fundamental theorem of algebra they cannot be equal. Hence this case is contradictory.
case 2:
p does not divide a
Now we have:
b^2mp^{2k+1}=a^2
Where the LHS is clearly divisible by p and the RHS is clearly not divisble by p so this case is also contradictory.

Since the assumption that \sqrt{n} can be written as a reduced fraction leads to contradictions, it must be incorrect.

Since \sqrt{n} cannot be written as a reduced fraction, it cannot be written as a fraction, so it must be irrational. QED

 

[modmans2ndcoming]

actualy, it seemed to me that the OP was trying to proove that SQRT of 6 is irrational.

the OP did not say if they had prooven that root 2 and root 3 were irrational.

 

[modmans2ndcoming]

yes, proof by contradiction is probably the best meathod for this since it is so short and uncomplicated.

 

[markgriffith]

Am I missing something?

Why is it not enough to note that root 6 divides by root 2?

.

 

[Muzza]

sqrt(6) "divides by" sqrt(2)? What does that mean?

 

[selfAdjoint]

It means \sqrt 6 = \sqrt 2 \sqrt 3. But how can this be used to show \sqrt 6 is irrational? Do you have a theorem that the product of irrationals is irrational? Surely not, since \sqrt 2 \sqrt 2 = 2.

 

[robert Ihnot]

These problems beginning with the sqr(2) can all be approached the same way if they are not perfect squares. We set p/q = sqr(6). Where p and q are integers without a common term. Then we have p^2 = 6q^2.

Now we have a theorem that if a prime u divides axb than p divides a or it divides b.

Since 2 and 3 are primes, they both divide p giving p=6k.

Then going back to the original problem we get: 36k^2 = 6q^2.

Thus we find that 6 = 2x3 now divides q. But this is imossible since p and q were taken in their smallest terms, where they have no common factor. QED

 

[markgriffith]

the cases when two irrational numbers have an irrational product

.

I'm probably being totally naive, but I'd have thought it is not too hard to show that two irrational numbers (a, b) have an irrational product if a and b are not equal, and a is not the reciprocal of b.

Not straightforward to show?

Mark

.

 

[Gokul43201]

Mark Griffith,
read through previous posts - this have been covered before. If you're still not happy, multiply sqrt(6) and sqrt(24).

 

[geraldmcgarvey]

sqrt(6) is irrational

Suppose sqrt(6) = a/b and a/b is in reduced form, i.e. a and b are relatively coprime. Then 6b^2 = a^2. Let x|y denote 'x divides y'.
2|6b^2 and 3|6b^2
2|a^2 and 3|a^2
2|a and 3|a
6|a
6^2|a^2
6^2|6b^2
6|b
So 6|a and 6|b which contradicts the assumption that a and b are relatively coprime. Therefore sqrt(6) is irrational.

 

[pmoseman]

Mark Griffith,
read through previous posts - this have been covered before. If you're still not happy, multiply sqrt(6) and sqrt(24).

Obviously sqrt(24) is just sqrt(6) * 2 so you haven't proven anything but 6 * 2 = 12.

 

[HallsofIvy]

Mark Griffith,
read through previous posts - this have been covered before. If you're still not happy, multiply sqrt(6) and sqrt(24).

Obviously sqrt(24) is just sqrt(6) * 2 so you haven't proven anything but 6 * 2 = 12.

Gokul's post was, as he said, in response to

.

I'm probably being totally naive, but I'd have thought it is not too hard to show that two irrational numbers (a, b) have an irrational product if a and b are not equal, and a is not the reciprocal of b.

Not straightforward to show?

Mark

.

\sqrt{6} and \sqrt{24} and irrational, not equal and not reciprocals.
\sqrt{6}\sqrt{24}= \sqrt{144}= 12, a rational number.

Which proves that, far from being "straightforward to prove", it is not true!

 

[pmoseman]

Square roots of irrational numbers have how many fewer rational solutions?

Look at the relationships of square roots laid out with respect to one another. Please take a moment to warm your brain up. This is not a proof and debate is welcome.

In the following equations:
(1) sqrt(X) + sqrt(Y)
(2) sqrt(X) - sqrt(Y)
(3) sqrt(X) / sqrt(Y)
(4) sqrt(X) * sqrt(Y)

Unless X and Y are both perfect squares, one will form an irrational term in the equation.

To obtain a rational solution, when one term is irrational, both terms must be irrational.

Range of rational solutions with irrational terms:
(1) NON
{}

(2) 0
{0}

(3) positive rational numbers and 0
{0, |Q|}

(4) If X or Y are themselves irrational the only solution is 0.
When X and Y are rational the solution includes 0 and all positive rational numbers.
{0} or {0, |Q|}

Range of rational solutions with rational terms:
(1) natural numbers and 0
{0, 1, 2, 3, 4, ...}

(2) integers
(...3, 2, 1, 0 ,-1, -2, -3, ...)

(3) positve rational numbers and 0
{0, |Q|}

(4) natural numbers and 0
{0, |Q|}

Square-roots of perfect squares are natural numbers.

All bases X and Y have one divisor which is a perfect square (1, 4, 9, 16, 25...).

When these equations have rational solutions, and X and Y have been reduced until the only divisor from the perfect squares is 1, then they are equal [rX = rY].

The natural number (square-root of the perfect square divisor) is distributed to the coefficient.

(3) sqrt(rX) / sqrt(rY) = 1, leaving the division of coefficients as a solution.
(4) sqrt(rX) * sqrt(rY) = rY which is multiplied by coefficients. This rY is some rational number.

My question, is... in the case of X and Y being themselves irrational, do you see other reasoning, perhaps, to test the irrational solutions? ln(pi)

Is everything above true, if so is it true for all cases of irrational terms, even those formed by irrational bases?

 

[pmoseman]

I'm probably being totally naive, but I'd have thought it is not too hard to show that two irrational numbers (a, b) have an irrational product if a and b are not equal, and a is not the reciprocal of b.

Not straightforward to show?

Mark

Looking at this quote, I think we can keep along the same line of thought as Mark was having, and say that two irrational numbers must have an irrational product under certain circumstances.

I believe it is safe to say, on Mark's behalf here, that a rational and irrational product can never be rational, just as an even and odd product can never be odd. In my book, irrational numbers seem like the "odd" ones.

Do all irrational products have rational solutions? Clearly not by the generalized proof of square roots of non-perfect square numbers.

Do some irrational products have rational solutions, yes, but the solutions of rational terms are limited only to rational products. So how do we pass through a door one way and are unable to pass through the other way? We must have reached some stability with rational numbers while irrational numbers remain balanced in a precarious way, making them unpleasant.

Multiplying an irrational number by a rational number is tantamount to coloring the roses red, when they are actually white. Sure you can show change an irrational number and show they multiply, but deep down somewhere unique to that number is a resonance that says, I AM THE SQRT(2).

What would be interesting, what I am getting back to with the above information is can you show me a case where rY/rX = rZ among distinct irrational numbers? Where rZ is not just some infinite non-repeating random decimal but actually has a solution we can muster in rational terms?

Can irrational numbers be reciprocals? The reciprocal of sqrt(2) is sqrt(.5), dividing these you get sqrt(4). I see that, I know that.

rY/rX =rZ in irrational terms sqrt(rY)/sqrt(rX) = sqrt(rY/rX) so rY/rX = rZ. But now, what if rY and rZ are themselves irrational? Like rY = sqrt(2) and rX = sqrt(5) then sqrt(rY)/sqrt(rX) = sqrt(rZ) and rZ = sqrt(2) / sqrt(5) so sqrt(sqrt(2/5)) is the final solution

Can you express the reciprocal of sqrt(2) for instance without giving me the trivial solution of 2^-0.5 or 0.5^1/2?

These numbers are simply reciprocals by definition of the word reciprocal. I think most people look at this topic from a deeper level, words like irrational, random, infinite conjure up a host of possibilities.

blargh!

 

[pmoseman]

And then prove the theorem:
Any number whose prime factorization has any odd exponents has an irrational square root.

I am having trouble for the moment understanding this theorem, and I have a question about this theorem in general, even if it is in fact a theorem.

What does prime factorization have to do with even/odd exponents? Prime factorization leads to a set of prime numbers. Can a prime number have any exponent except for 1? This theorem says all numbers have an irrational square root.

A theorem should be able to stand alone, and this theorem doesn't seem to make sense. You mean a number "has" an irrational square root. It would be clearer to state, "The square root of any number whose prime facorization has an odd exponent is irrational."

 

[Gokul43201]

pmoseman: You are using some very non-standard terminology that makes it hard to understand what you're saying. For instance, the things that you refer to as equations (in post #45) are actually expressions. Equations have solutions, expressions have values.

 

[Gokul43201]

I am having trouble for the moment understanding this theorem, and I have a question about this theorem in general, even if it is in fact a theorem.

What does prime factorization have to do with even/odd exponents? Prime factorization leads to a set of prime numbers. Can a prime number have any exponent except for 1? This theorem says all numbers have an irrational square root.

A theorem should be able to stand alone, and this theorem doesn't seem to make sense. You mean a number "has" an irrational square root. It would be clearer to state, "The square root of any number whose prime facorization has an odd exponent is irrational."

Any number, n, can be expressed as a product of its prime factors: n = p_1^a \cdot p_2^b \cdots p_k^m (example: 12=2²3¹). The numbers a,b,...,m are the exponents in the prime factorization. In the above example (n=12) there is one odd exponent. Clear?

 

[pmoseman]

pmoseman: You are using some very non-standard terminology that makes it hard to understand what you're saying. For instance, the things that you refer to as equations (in post #45) are actually expressions. Equations have solutions, expressions have values.

Are you calling me out on using the word "equations", instead of "expressions" (for which I provide solutions).
It seems lackadaisical then for you to use terms such as "any" odd exponent, or a number "has" a square root. I could only understand the theorem you stated after seeing this: n = p_1^a \cdot p_2^b \cdots p_k^m (example: 12=2²3¹).

It can be stated clearly:
"The square root of a number is irrational, if the complete prime factorization of that number includes a prime any odd number of times."

This rings true; "prove it" seems a bit of a challenge. It should work for the square root of any rational number.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I know the list of solutions to (1) (2) (3) (4) is probably hard to understand, but I simply did that to make a point.

I do not know of any terminology to express X and Y reduced by the largest perfect square divisor, so I simply wrote rX and rY (reduced X and reduced Y). I think it is pretty clear/simple, since factoring out perfect squares "reduces" a square root. sqrt(Y) = k*sqrt(rY), where k is a natural number equal to the square root of the largest prime divisor of Y.

If there is anything else that confuses you, I would want myself to be understood.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Mark simply asked if it would be straighforward "...to show that two irrational numbers (a, b) have an irrational product if a and b are not equal, and a is not the reciprocal of b."

I feel like I have shown Mark's naive idea can be eked into a straightforward theorem.
"Two irrational numbers (X, Y) have an irrational product if rX and rY are not equal, and rX is not the reciprocal of rY." In fact, the only rational solution to most any of it, is 0, and 0 is a "rational" number only by acquaintance. In the most general way Mark was right on the money, although he might not be majoring in mathematics.

I mean, honestly, there is no justification to the approach of disproving his question by showing sqrt(24) * sqrt(6) = 12 because sqrt(24) and sqrt(6) are not really different irrational numbers, in the way 4 and 2 are not actually different numbers, they are different amounts of the same prime number, paying no mind to the number 1. He shouldn't have to worry about them being reciprocals either, like Matt Grime's example of pi *1/pi, since that would technically make a quotient, but alas. Are we going to clap Matt on the back for proving pi and pi^-1 are "different" irrationals that "produce" 1?

In fact, doesn't Mark's line of thinking help us prove that sqrt(6) is irrational? It is closely related to the many approaches presented to this problem.

By proving sqrt(3) and sqrt(2) are not equal when reduced, we could then be able to show, with such an amazing theorem, their product is an irrational number. So thanks Mark.