- #36
markgriffith
- 10
- 0
Am I missing something?
Why is it not enough to note that root 6 divides by root 2?
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Why is it not enough to note that root 6 divides by root 2?
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Gokul43201 said:Mark Griffith,
read through previous posts - this have been covered before. If you're still not happy, multiply sqrt(6) and sqrt(24).
Gokul43201 said:Mark Griffith,
read through previous posts - this have been covered before. If you're still not happy, multiply sqrt(6) and sqrt(24).
pmoseman said:Obviously sqrt(24) is just sqrt(6) * 2 so you haven't proven anything but 6 * 2 = 12.
[itex]\sqrt{6}[/itex] and [itex]\sqrt{24}[/itex] and irrational, not equal and not reciprocals.markgriffith said:.
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
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NateTG said:And then prove the theorem:
Any number whose prime factorization has any odd exponents has an irrational square root.
Any number, n, can be expressed as a product of its prime factors: [itex]n = p_1^a \cdot p_2^b \cdots p_k^m[/itex] (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 said: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 said: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.
TylerH said:Is it enough, once given that, if a|bn, then an|b[supn[/sup], to say that:
n1/2=p/q
n=p2/q2
nq2=p2 Contradiction if n is not expressible as k2, 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.
You don't have to at all.pmoseman said: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.
I'm saying your posts are more likely to be understood if you use standard terminology. If you don't care to have your posts understood, by all means, call an expression a banana.pmoseman said:Are you calling me out on using the word "equations", instead of "expressions" (for which I provide solutions, so wtfc?)?
I never used any such terms. But surely, you understand the difference between using inelegant English and wrong terminology. Not everyone here is a native speaker of English.It seems lackadaisical then for you to use terms such as "any" odd exponent, or a number "has" a square root.
Sucks to be me, for taking the trouble of explaining a theorem to you (that someone else stated).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=2231).
Kudos to your English teachers.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."
If the list is hard to understand, then your point was probably missed.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.
2 and 4 are not different numbers? Neither are sqrt(24) and sqrt(6)? That requires a serious twisting of either language or terminology.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.
Like it or not, every number other than the multiplicative identity is indeed most definitely different from its inverse. You can not choose to use some non-standard interpretation of the word 'different' to argue that all of here are wrong.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?
Gib Z said:I wonder why this proof is not considered more standard:
Clearly the only integers whose square roots are integers are the perfect squares. So let's deal with a non perfect square, n. Suppose [tex] \sqrt{n} = \frac{p}{q} [/tex] where we can suppose gcd(p,q) = 1, and [itex] q> 1 [/itex]. Then [tex] n = \frac{p^2}{q^2} [/tex] which can't occur, since there are no common factors to cancel to get a denominator of 1 which is required to equal to integer, n. So we have a contradiction and square roots of non perfect squares are irrational.
An irrational number is a real number that cannot be expressed as a ratio of two integers. This means that the decimal representation of an irrational number is non-terminating and non-repeating.
We can prove that sqrt(6) is irrational using a proof by contradiction. Assume that sqrt(6) is rational, meaning it can be expressed as a ratio of two integers a/b where a and b have no common factors. Squaring both sides of this equation, we get 6 = a^2/b^2. This means that a^2 is divisible by 6, and therefore a must also be divisible by 6. However, if a is divisible by 6, then a^2 is divisible by 36. This means that b^2 must also be divisible by 36, which contradicts our assumption that a and b have no common factors. Therefore, our initial assumption that sqrt(6) is rational must be false, and thus it is irrational.
Yes, we can approximate sqrt(6) with a rational number by using a decimal approximation. For example, sqrt(6) is approximately 2.44948974278. However, this is only an approximation and not the exact value of sqrt(6).
Yes, there are other methods to prove that sqrt(6) is irrational. One method is to use the irrationality of sqrt(2) and the fact that the product of two irrational numbers is also irrational. Another method is to use the fact that the sum or difference of a rational and an irrational number is irrational. Both of these methods require some knowledge of advanced mathematical concepts.
The irrationality of sqrt(6) (and other numbers) has important implications in mathematics and other fields. It shows that there are numbers that cannot be expressed as a ratio of two integers, and this has implications in number theory, geometry, and other areas of mathematics. It also has practical applications in fields such as cryptography, where the use of irrational numbers makes encryption more secure.