| Posted on Sunday, 23 November, 2003 - 09:59 pm: |
Can someone tell me if i have proved the following validly or not? If not, the say so, but please don't tell me how to do the problem
Show that if p is prime, then p^1/2 must be irrational
p|ab => pn = ab => pc= a and pd = b
for n,c,d elements of Z and a,b elements of N
so (p^2)cd = ab
(p^2)cd = pn
p = n/cd
Now for a number to be rational, it must be able to be expressed in form s/t where s,t are elements of Z, without common factors.
so assume p^1/2 is rational
p^1/2 = s/t
p^1/2 = (n/cd)^1/2
where n = ab/p and cd= ab/p^2
and so n and cd have a common factor of ab which is a contradiction. Therefore the square root of any prime must be irrational
Thanks
Dean
| Posted on Sunday, 23 November, 2003 - 10:12 pm: |
Deleted
| Posted on Sunday, 23 November, 2003 - 10:47 pm: |
Unfortunately, there is a problem on line 1, since:
but
| Posted on Sunday, 23 November, 2003 - 10:50 pm: |
I dont get the last statement, can you expand. Is what your saying that since a rational number can be expressed as s/t where s and t are coprime, therefore its a contradiction since n and cd arent coprime?
If you sub back in you get
p = n/cd = ab/p / ab/p2
= ab/p * p2 /ab
= p
I dont really understand.
By the way, the fact that p is prime is not used in the proof and therefore if your proof is correct, it can prove that same result for any number, even if its not prime.
| Posted on Sunday, 23 November, 2003 - 10:56 pm: |
Heh, i just saw ur message Tristan, thats what i said before i deleted it.
The reason i deleted it is because if
p|ab, then either;
p|a or,
p|b, or
p|a and p|b
So he assumes the latter in his proof, i dont think thats a problem. However it is something to note Dean if you didnt know that already
| Posted on Sunday, 23 November, 2003 - 11:28 pm: |
It's a brave attempt, Dean, but you've used circular reasoning.
The first line is where your problems start. You can certainly say that if p|ab, then pn=ab. But you cannot maintain that pc=a and pd=b, where c and d are integer. For example, 5|6x10, 5x12=6x10, but 5x1.2=6 and 5x2=10.
If you start with the opposite (or assume that a and b are chosen such that), pc=a and pd=b, hence p|ab, then you have made p2 |ab, as ab=p2 cd.
I don't see where the problem with the final line comes from too. Just because s and t have no common factors, doesn't mean that the equality, s/t=n/cd, requires n and cd to have no common factors: 2/1=12/6=2. Why should cd and n have no common factors? In fact your fourth line: (p^2)cd = pn, shows that n|cd.
May I suggest you start by assuming that p1/2 =a/b, where HCF(a,b)=1.
| Posted on Monday, 24 November, 2003 - 11:41 am: |
Right ok, so my main problem is the fact that p|ab doesn't imply p|a AND p|b? Where will i find a contradiction then in order to prove what i want? I want basically to arrive at a contradiction in a similar way as in a proof that square root of 2 is irrational, but more generally for a prime. Any hints on how to do this?
Colin: is it implicit in your assumption that p is prime?
Thanks
Dean
| Posted on Monday, 24 November, 2003 - 12:24 pm: |
Dean,
Start with p1/2 =a/b and a,b have no common factors, and assume that p is prime. Try to argue that a must have a factor of p, and thence that b must have a factor of p, which contradicts the 'no common factor' assumption.
Andre
| Posted on Monday, 24 November, 2003 - 06:51 pm: |
Ok, i refined the proof to the following
Let P^1/2 = a/b where a and b have no common factor and p is prime then we have
p = (a/b)^2
pb^2 = a^2
which implies p|a^2
i.e. p|a x a, which implies p|a or p|a
Now since p|a then a= pc for some c
so pb^2 = a^2
pb^2=p^2.c^2
b^2 =pc^2
which implies p|b^2 which implies p|b
which shows that both b and a have a common factor which is a contradiction. Therefore p^1/2 must be irrational
Does this seem more valid?
Dean
| Posted on Monday, 24 November, 2003 - 07:07 pm: |
It looks good to me. You might also have a look here
Andre
| Posted on Monday, 24 November, 2003 - 07:24 pm: |
There's another nice proof, involving some analysis. We show that if x is an integer then x is a perfect square or sqrt(x) is irrational.
Suppose x is not a perfect square, so sqrt(x) is not an integer. Then there exists an integer k such that:
0 < sqrt(x) - k < 1
Consider raising both sides to the power n. By the binomial theorem (sqrt(x) - k)n = an sqrt(x) + bn for some integers an ,bn . Now if sqrt(x) is irrational then sqrt(x) = p/q with p,q integers and q > 0.
Then (sqrt(x) - k)n = (an p + bn q)/q. But the left hand side is positive. Therefore an p + bn q must be positive. But it is an integer, therefore it is > = 1.
Hence: (sqrt(x) - k)n > = 1/q for all n.
However since 0 < sqrt(x) - k < 1, (sqrt(x) - k)n -> 0 as n -> infinity which is a contradiction.
The same method can easily be extended to show that if P is a monic polynomial with integer coefficients, and P(x) = 0 then x is either an integer or irrational.
Michael
| Posted on Monday, 24 November, 2003 - 07:31 pm: |
Quote :
"Now if sqrt(x) is irrational then sqrt(x) = p/q with p,q integers and q > 0."
hmm it should be sqrt(x) is rational or sqrt(x) is not irrational.
( excuse my nitpicking, just trying to be overly smart)
love arun
| Posted on Monday, 24 November, 2003 - 08:01 pm: |
If a polynominal is monic, does it mean that the roots are all integers or all irrational, ie. theres not a mix? I presume this is what you mean but i just want to make it clear.
| Posted on Monday, 24 November, 2003 - 08:07 pm: |
Well firstly, monic just means that the coefficient of the highest power of x is 1. If the other coefficients are also integers, then the roots are either integers or irrationals, or a mixture of both. As an example of the latter, consider expanding out: (x-1)(x-Ö2)(x+Ö2)=0 Andre
| Posted on Monday, 24 November, 2003 - 08:26 pm: |
Oh i see now. I was being a bit stupid...I forgot about the existance of non-integer rational numbers.
Thanks
| Posted on Tuesday, 25 November, 2003 - 06:19 am: |
You can also do it using the fundamental theorem of arithmetic. We aim to show Öx is either an integer or irrational. If x is a perfect square then its square root is (by definition) and integer. So, suppose x isn't a perfect square. Thus it must contain at least one prime factor an odd number of times, call it p. Now, assume Öx=a/b where a and b are integers. Therefore a2=x b2. But, p will appear in both a2 and b2 an even number of times, and hence will appear an odd number of times in x b2. By the FTA, this is a contradiction.
(I hope I haven't assumed anything I wasn't meant to in that proof)
Marcos
| Posted on Tuesday, 25 November, 2003 - 06:21 am: |
I've just noticed that this is practically the same as Andre's and mine can be extended in the same way for monic polynomials... Oh well...
| Posted on Tuesday, 25 November, 2003 - 07:33 pm: |
Apart from Michael's proof, and the one based on the Fundamental Theorem of Arithmetic, the irrationality of Ön (if not an integer) can also be derived through continued fractions: (i) Show that the continued fraction of a positive real is unique (ii) Show that the continued fraction expansion for a/b (where a, b are positive integers) terminates (iii) Show that the continued fraction for Ön does not terminate (in fact it is periodic). Thus Ön cannot be equal to a/b. Of the three proofs, this one is probably the most trouble, however there are additional prizes to be hard. For example, we can write e as a simple continued fraction (where the constants follow the pattern [1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, ..]). This immediately shows e to be irrational. Andre
| Posted on Wednesday, 26 November, 2003 - 02:00 pm: |
Thanks for the correction Arun. Also: I said "consider raising both sides to the power n" when I really meant just "consider raising sqrt(x)-k to the power n".
| Posted on Wednesday, 26 November, 2003 - 06:30 pm: |
Andre,
how does one start with (iii)?
Michael,
i understand. it pretty much happens with me always. In some earlier post i typed in "thinks" instead of "things".(dunno what i was "think"ing then)
love arun
| Posted on Thursday, 27 November, 2003 - 11:53 am: |
Arun,
According to Davenport (The Higher Arithmetic), the theorem that "any quadratic irrational number has a continued fraction which is periodic after a certain stage" was first proved by Lagrange in 1700. You can find a proof in either Davenport or Hardy (Introduction to the theory of numbers). I can write it out if you wish but it's a bit long ...
Andre
| Posted on Thursday, 27 November, 2003 - 12:02 pm: |
Andre,
if its long then i will ask this problem after i am through with my exams. I don't think i am in any mood to understand big long proofs during my EPL(exam preparation leave).
Anyways, Thanks!
love arun