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anonymous
 one year ago
(Introductory Real Analysis) I'm trying to show that sqrt(p) is irrational if p is prime, but I have no idea where to start; could anybody give me suggestions without outright telling me what to do? e.g. should I start from a specific type of argument, contrapositive/counterexample/contradiction, etc, or should I mess around with some field axioms?
anonymous
 one year ago
(Introductory Real Analysis) I'm trying to show that sqrt(p) is irrational if p is prime, but I have no idea where to start; could anybody give me suggestions without outright telling me what to do? e.g. should I start from a specific type of argument, contrapositive/counterexample/contradiction, etc, or should I mess around with some field axioms?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0contradiction would be your best bet i reckon? dan is the master so listen to him

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i really dont know what to do yet xD

dan815
 one year ago
Best ResponseYou've already chosen the best response.2but im thinking contradiction is the easiest too

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's what I'm thinking too, messing around with it on a whiteboard. Prime numbers cannot be even, so p must be odd; both a and b must be either both odd or both even because of p being odd.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now I'm thinking of trying to prove a contradiction from this, or something. I'll see in a minute.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A and b can't be even, because they wouldn't be in lowest terms (where a and b are the integers forming a fraction in lowest terms, equaling sqrt(p).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So a and b must both be odd. Now I'll try to contrive a contradiction from this, but IDK if this will work the same way as it does with something like sqrt(3).

dan815
 one year ago
Best ResponseYou've already chosen the best response.2now lets say sqrt(p) is a rational then there is some lower fraction m/n where n doesnt divide m

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, what is your contradiction?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your picture made zero sense to me.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2then if the sqrt(p) = m/n then p=m^2/n^2 n^2*p=m^2 now we know that n doesnt divide m? that means that p divide m^2 but if p is a factor of m^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"now we know that n doesnt divide m?" How do we know that?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2then that means in m^2 there has to be a p^2

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh because remember than every rational number can be reduced to some lowest fraction

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i know theres definately some contradiction about this lemme thing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah.....\[3c^2=d^2\]Both c^2 and 3 must be a factor of d^2. c^2 is equal to an integer, and d^2 is equal to an integer; c^2 is divisible by d^2 by that relation, that isn't the contradiction, or at least in every other proof regarding sqrt(2), sqrt(3) and so on, I have never ever seen that used as the contradiction.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2n^2*p=m^2 n^2*p = k^2*p^2 n^2=k^2*p but n^2 cannot be a multiple of p because both n doesnt divide m

dan815
 one year ago
Best ResponseYou've already chosen the best response.2thats a contradiction there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You know, LaTeX makes it way clea...yeah, that's better.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2okay LETS write this out agin clearly now that we know

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here is what i gathered up from 1st yr notes if this helps consolidate what dan is saying

dan815
 one year ago
Best ResponseYou've already chosen the best response.2suppose \[\sqrt(p)=m/n\\ p=m^2/n^2\\ n^2*p=m^2\\ n^2*p=k^2*p^2\\ n^2=k^2*p\\\] but the last step means the same as the 3rd last step, that n^2 is a multiple of p^2 that means n would divide m and by definition lowest term fraction wont do that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm literally just not sure where you got this line here, \[n^2p=k^2p^2.\] You're saying m is equal to kp and then squaring it? On what grounds? From this relation:\[n^2p=m^2\]I can say that p is a factor of m^2, and n^2 is also a factor of m^2, but I haven't proven or otherwise demonstrated that p is a factor of m, have I?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2if p is a factor of m^2 then u must have p^2 in m^2

dan815
 one year ago
Best ResponseYou've already chosen the best response.2other wise its not possible

dan815
 one year ago
Best ResponseYou've already chosen the best response.2for any square integer there have to be atleast 2 of every prime

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, that's something that needs to be proven or otherwise justified. That's not a field axiom or anything.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2since any given integer = p1*p2*p3... so integer^2 = p1^2*p2^2... atleast it can be that the integer pas p1^2 in it already and it would become p1^4

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh im not sure about axioms and stuff

dan815
 one year ago
Best ResponseYou've already chosen the best response.2the only thing i know is the basic principle that all integers are products of primes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm still confused about you saying that m^2 is not divisible by p. In the explanation @chris00 posted regarding sqrt(3), it makes the same relation but says the exact opposite, that its version of m^2 must be divisible by 3.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The relation \[pn^2=m^2\]Would imply to me that m^2 must be divisible by p. Why is this not correct.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2http://prntscr.com/8cr1bj do u understand this conclusion

dan815
 one year ago
Best ResponseYou've already chosen the best response.2how that certain statement has to gurantee an existance of some factor * p^2 in it

dan815
 one year ago
Best ResponseYou've already chosen the best response.2since m and n are integers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where are these additional variables coming from? I'm now more confused than before, what the heck is the point of introducing k_1 and k_2?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2what we basically established is that m^2 and n^2 are factors of p but n doesnt divide m neither does n^2 divide m^2 then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol. Suddenly random variables.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2im just saying it has to be some otehr factor of p,

dan815
 one year ago
Best ResponseYou've already chosen the best response.2let k1^2 and k2^2 be some random factors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[p=\frac{m^2}{n^2} \ \ \ \text{Agreed.}\]\[pn^2=m^2 \ \ \ \text{Yup.}\] \[\text{m^2 is equal to n^2p, so p must be a factor of m^2.}\]\[m^2=pk.\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2tell also k happens to be a perfect square

dan815
 one year ago
Best ResponseYou've already chosen the best response.2p^2 must be a factor of m^2 actually

dan815
 one year ago
Best ResponseYou've already chosen the best response.2okay lets work from here , this seems to be the most confusing part p*n^2=m^2

dan815
 one year ago
Best ResponseYou've already chosen the best response.2now lets start with the basic definition of m

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"tell also k happens", wot. Dan, you are possibly the best mathematician on OpenStudy who isn't a straight up university professor (yet), but I have a super hard time understanding your worded explanations, lol. Alright, let's work from there, yeah.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2since m is an integer it can always be written has some product of primes

dan815
 one year ago
Best ResponseYou've already chosen the best response.2now p*n^2=m^2 okay now if p divides m^2 and since its m^2 every prime in m^2 has a multiplicity of atleast 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have no idea what the tiny zero things are that somewhat resemble subscripts but wouldn't make any sense being subscripts, what those are in the picture.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2every integer is a product of primes is what im saying

dan815
 one year ago
Best ResponseYou've already chosen the best response.2every integer product of primes with some exponent

dan815
 one year ago
Best ResponseYou've already chosen the best response.2for example 25=5^2 15=3*5 15^2=3^2*5^2

dan815
 one year ago
Best ResponseYou've already chosen the best response.2notice how every squard integer will have atleast 2 exponent on every prime

dan815
 one year ago
Best ResponseYou've already chosen the best response.2its always a multiple of 2

dan815
 one year ago
Best ResponseYou've already chosen the best response.2are u good with this step now?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2p*n^2=m^2 then p*n^2=p^2*k^2

dan815
 one year ago
Best ResponseYou've already chosen the best response.2heres some examples to help u see clearly 3*k = 15^2 if 3 is a factor of 15^2 then there must be atleast 3^2 in 15^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, I think I get your reasoning underlying this specifically, wait a minute.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If we were dealing with a^2 equal to k times any given value, my retort to what you originally said, that p must be squared, would be, what if p as a factor was the square of a number already? Since p is stated as a prime (that's the thing that somewhat sets this apart from other possible sqrt irrationality proofs), we know that it can't be further simplified in any way, thus p can't be equal to, say, the square of a prime digit; it has to be equal to that prime digit itself.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So that justifies m^2=p^2k^2; if p is prime and it is a factor of m^2, p^2 is also a factor of m^2.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, cool. Agreed.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2okay now ther is an intermediate step i didnt state cuz its kidn of implied

dan815
 one year ago
Best ResponseYou've already chosen the best response.2p*n^2=m^2 now we agree p*n^2=p^2*k^2 this means for these to be equal the other p must exist in n^2 but if that happened the n would have divided m, which is a contradiction

dan815
 one year ago
Best ResponseYou've already chosen the best response.2what we did earliier was very similiar but turns out we could have stopped at this step itself

dan815
 one year ago
Best ResponseYou've already chosen the best response.2what we did ealier was p*n^2=m^2 p*n^2=p^2*k^2 divide both sides by p n^2=p*k^2 which is an equilvalent statement to the first statement meaning n^2=k2^2*p^2, that n^2 is some other factor of p^2 so if n^2 and m^2 were both factors of p^2 then n would have divided m and another contradiction arises

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I agree with your first explanation of the contradiction where m divides n, or otherwise stated that both m and n were note in simplest terms and thus p cannot be rational, but I don't understand this part of the last post:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"which is an equivalent statement to the first statement"; Wait, so you're saying \[n^2=pk^2 \leftarrow \rightarrow np^2=m^2\] Is this what you mean by "first statement"?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the "first statement" exactly, lol.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so the same reason we concluded m^2 is a factor of p^2 then n^2 is a factor of p^2 because of that similiar expression arising

dan815
 one year ago
Best ResponseYou've already chosen the best response.2theres actually another contradiction argument for completeness sake

dan815
 one year ago
Best ResponseYou've already chosen the best response.2if p*n^2=p^2*k^2 then there are an even exponent of p on the right and an off expoenent for p on the left

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so regardless of that fact that n^2 and k^2 might be factors of p, this can never be equal as there will always be odd p exponent on and even p exponent on either side

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I understand the first and third explanations of contradiction, I don't understand the second one, heh. I'll try to have it explained by someone else because I'm just having trouble understanding your words, I'm sure it makes perfect sense to most people but I have an unreasonably difficult time understanding you sometimes, lol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And thank you so much for your help, you're really good at this.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2like okay we saw that for ever square all the exonents must be factors of 2

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so for p*n^2 = p^2*k^2 so lets say even if there is a p hidden in n^2 or a p hidden in k^2 they will have even expoennt

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so the left side will always have odd* exponent for p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I understand the third one, I didn't understand the second one. The exponent contradiction makes sense to me, and the first one does, too, I said it was the middle one.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2oh which 2nd one you mean the one in the screenshot?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2so if \[n^2=p*k_1^2\] then \[n^2=p^2*k_2^2\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2just like how \[m^2=p*n^2\\ \] \[m^2=p^2*k^2\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.2both cases we are showing that n^2 and m^2 had a p^2 in therm which cannot be

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, think I get it. I think repetition of working through the proofs for this will help. It might just take time to get used to it. I'm still weirded out by the fact that the set of rational numbers is countable.
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