Mendicant_Bias
  • Mendicant_Bias
(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?
Mathematics
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Mendicant_Bias
  • Mendicant_Bias
(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?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
contradiction would be your best bet i reckon? dan is the master so listen to him
dan815
  • dan815
i really dont know what to do yet xD
dan815
  • dan815
but im thinking contradiction is the easiest too

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Mendicant_Bias
  • Mendicant_Bias
That'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.
Mendicant_Bias
  • Mendicant_Bias
Now I'm thinking of trying to prove a contradiction from this, or something. I'll see in a minute.
Mendicant_Bias
  • Mendicant_Bias
A 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).
Mendicant_Bias
  • Mendicant_Bias
So 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
  • dan815
|dw:1441420377401:dw|
dan815
  • dan815
i think this works
dan815
  • dan815
now lets say sqrt(p) is a rational then there is some lower fraction m/n where n doesnt divide m
Mendicant_Bias
  • Mendicant_Bias
Wait, what is your contradiction?
Mendicant_Bias
  • Mendicant_Bias
Your picture made zero sense to me.
dan815
  • dan815
then 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
Mendicant_Bias
  • Mendicant_Bias
"now we know that n doesnt divide m?" How do we know that?
dan815
  • dan815
then that means in m^2 there has to be a p^2
dan815
  • dan815
oh because remember than every rational number can be reduced to some lowest fraction
dan815
  • dan815
i know theres definately some contradiction about this lemme thing
dan815
  • dan815
think*
Mendicant_Bias
  • Mendicant_Bias
Yeah.....\[3c^2=d^2\]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.
anonymous
  • anonymous
ye nah ur right dan
dan815
  • dan815
n^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
  • dan815
thats a contradiction there
Mendicant_Bias
  • Mendicant_Bias
You know, LaTeX makes it way clea-...yeah, that's better.
dan815
  • dan815
okay LETS write this out agin clearly now that we know
anonymous
  • anonymous
here is what i gathered up from 1st yr notes if this helps consolidate what dan is saying
dan815
  • dan815
suppose \[\sqrt(p)=m/n\\ p=m^2/n^2\\ n^2*p=m^2\\ n^2*p=k^2*p^2\\ n^2=k^2*p\\\] 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
  • anonymous
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Mendicant_Bias
  • Mendicant_Bias
I'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
  • dan815
oh because
dan815
  • dan815
if p is a factor of m^2 then u must have p^2 in m^2
dan815
  • dan815
other wise its not possible
dan815
  • dan815
for any square integer there have to be atleast 2 of every prime
Mendicant_Bias
  • Mendicant_Bias
Alright, that's something that needs to be proven or otherwise justified. That's not a field axiom or anything.
dan815
  • dan815
since 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
  • dan815
oh im not sure about axioms and stuff
dan815
  • dan815
the only thing i know is the basic principle that all integers are products of primes
Mendicant_Bias
  • Mendicant_Bias
I'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.
Mendicant_Bias
  • Mendicant_Bias
The relation \[pn^2=m^2\]Would imply to me that m^2 must be divisible by p. Why is this not correct.
dan815
  • dan815
http://prntscr.com/8cr1bj do u understand this conclusion
dan815
  • dan815
how that certain statement has to gurantee an existance of some factor * p^2 in it
dan815
  • dan815
since m and n are integers
Mendicant_Bias
  • Mendicant_Bias
Where 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
  • dan815
what 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
Mendicant_Bias
  • Mendicant_Bias
Lol. Suddenly random variables.
dan815
  • dan815
im just saying it has to be some otehr factor of p,
dan815
  • dan815
let k1^2 and k2^2 be some random factors
Mendicant_Bias
  • Mendicant_Bias
\[p=\frac{m^2}{n^2} \ \ \ \text{Agreed.}\]\[pn^2=m^2 \ \ \ \text{Yup.}\] \[\text{m^2 is equal to n^2p, so p must be a factor of m^2.}\]\[m^2=pk.\]
dan815
  • dan815
tell also k happens to be a perfect square
dan815
  • dan815
p^2 must be a factor of m^2 actually
dan815
  • dan815
but yes p too
dan815
  • dan815
okay lets work from here , this seems to be the most confusing part p*n^2=m^2
dan815
  • dan815
now lets start with the basic definition of m
Mendicant_Bias
  • Mendicant_Bias
"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
  • dan815
|dw:1441422178020:dw|
dan815
  • dan815
since m is an integer it can always be written has some product of primes
dan815
  • dan815
now 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
dan815
  • dan815
|dw:1441422303581:dw|
Mendicant_Bias
  • Mendicant_Bias
I 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
  • dan815
that is multiplication
dan815
  • dan815
every integer is a product of primes is what im saying
dan815
  • dan815
every integer product of primes with some exponent
dan815
  • dan815
for example 25=5^2 15=3*5 15^2=3^2*5^2
dan815
  • dan815
notice how every squard integer will have atleast 2 exponent on every prime
dan815
  • dan815
its always a multiple of 2
dan815
  • dan815
are u good with this step now?
dan815
  • dan815
p*n^2=m^2 then p*n^2=p^2*k^2
dan815
  • dan815
heres 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
Mendicant_Bias
  • Mendicant_Bias
Alright, I think I get your reasoning underlying this specifically, wait a minute.
Mendicant_Bias
  • Mendicant_Bias
If 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.
Mendicant_Bias
  • Mendicant_Bias
So 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.
dan815
  • dan815
yeah
Mendicant_Bias
  • Mendicant_Bias
Alright, cool. Agreed.
dan815
  • dan815
okay now ther is an intermediate step i didnt state cuz its kidn of implied
dan815
  • dan815
p*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
  • dan815
what we did earliier was very similiar but turns out we could have stopped at this step itself
dan815
  • dan815
what 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
Mendicant_Bias
  • Mendicant_Bias
I 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:
Mendicant_Bias
  • Mendicant_Bias
"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"?
Mendicant_Bias
  • Mendicant_Bias
What is the "first statement" exactly, lol.
dan815
  • dan815
http://prntscr.com/8cr8qz
dan815
  • dan815
so 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
  • dan815
any questios?
dan815
  • dan815
theres actually another contradiction argument for completeness sake
dan815
  • dan815
if 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
  • dan815
so 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
Mendicant_Bias
  • Mendicant_Bias
I 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.
Mendicant_Bias
  • Mendicant_Bias
And thank you so much for your help, you're really good at this.
dan815
  • dan815
like okay we saw that for ever square all the exonents must be factors of 2
Mendicant_Bias
  • Mendicant_Bias
Yeah, agreed.
dan815
  • dan815
so 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
  • dan815
so the left side will always have odd* exponent for p
Mendicant_Bias
  • Mendicant_Bias
Yeah, 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
  • dan815
oh which 2nd one you mean the one in the screenshot?
dan815
  • dan815
http://prntscr.com/8cr8qz
Mendicant_Bias
  • Mendicant_Bias
Yeah
dan815
  • dan815
so if \[n^2=p*k_1^2\] then \[n^2=p^2*k_2^2\]
dan815
  • dan815
just like how \[m^2=p*n^2\\ \] \[m^2=p^2*k^2\]
dan815
  • dan815
both cases we are showing that n^2 and m^2 had a p^2 in therm which cannot be
Mendicant_Bias
  • Mendicant_Bias
Alright, 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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