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Prove sqrt(6) is irrational 
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#37
May3004, 08:01 AM

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



#38
May3004, 02:46 PM

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



#39
May3004, 09:47 PM

P: 1,059

These problems begining 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 


#40
Jun104, 02:11 PM

P: 10

.
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 . 


#41
Jun104, 06:36 PM

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


#42
Jun1804, 10:57 PM

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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 xy denote 'x divides y'.
26b^2 and 36b^2 2a^2 and 3a^2 2a and 3a 6a 6^2a^2 6^26b^2 6b So 6a and 6b which contradicts the assumption that a and b are relatively coprime. Therefore sqrt(6) is irrational. 


#43
Feb1911, 04:13 AM

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#44
Feb1911, 11:06 AM

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[itex]\sqrt{6}\sqrt{24}= \sqrt{144}= 12[/itex], a rational number. Which proves that, far from being "straightforward to prove", it is not true! 


#45
Feb1911, 09:33 PM

P: 14

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} Squareroots 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 (squareroot 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? 


#46
Feb1911, 10:04 PM

P: 14

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 nonperfect 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 nonrepeating 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! 


#47
Feb2011, 06:21 PM

P: 14

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." 


#48
Feb2011, 08:50 PM

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pmoseman: You are using some very nonstandard 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.



#49
Feb2011, 08:56 PM

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#50
Feb2111, 04:37 PM

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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: [itex]n = p_1^a \cdot p_2^b \cdots p_k^m[/itex] (example: 12=2^{2}3^{1}). 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. 


#51
Feb2111, 09:15 PM

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It is not difficult to prove for arbitrary products of different primes (of course this will suffice for the general nonsquare case). The proof goes exactly the same way as the case for one prime. If [tex]b^2p_1...p_k = a^2[/tex], then [tex]p_1...p_k[/tex] must be a factor of a, etc..



#52
Feb2111, 09:52 PM

P: 737

Is it enough, once given that, if ab^{n}, then a^{n}b[supn[/sup], to say that:
n^{1/2}=p/q n=p^{2}/q^{2} nq^{2}=p^{2} Contradiction if n is not expressible as k^{2}, where k is an arbitrary integer. The contradiction being that there is an uneven power of a prime factor on one side, which is impossible for a squared number. 


#53
Feb2111, 10:08 PM

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This method of proof generalizes neatly to the n'th root case. Comparing the exponents of the primes in the respective factorizations will yield the wanted contradiction. 


#54
Feb2111, 11:57 PM

P: 14

How many times do we have to answer this guys question? Enough already!
We are not trying to do this to the nth root. Unless you have a proof other than sqrt(a) = p/q just forget it. That's the only proof anyone has. The rest are just statements of fact. 


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