A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Mendicant_Bias

  • 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?

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    contradiction would be your best bet i reckon? dan is the master so listen to him

  2. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i really dont know what to do yet xD

  3. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    but im thinking contradiction is the easiest too

  4. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  5. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Now I'm thinking of trying to prove a contradiction from this, or something. I'll see in a minute.

  6. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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).

  7. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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).

  8. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1441420377401:dw|

  9. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i think this works

  10. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    now lets say sqrt(p) is a rational then there is some lower fraction m/n where n doesnt divide m

  11. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wait, what is your contradiction?

  12. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Your picture made zero sense to me.

  13. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  14. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    "now we know that n doesnt divide m?" How do we know that?

  15. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    then that means in m^2 there has to be a p^2

  16. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh because remember than every rational number can be reduced to some lowest fraction

  17. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i know theres definately some contradiction about this lemme thing

  18. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    think*

  19. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ye nah ur right dan

  21. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  22. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    thats a contradiction there

  23. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You know, LaTeX makes it way clea-...yeah, that's better.

  24. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    okay LETS write this out agin clearly now that we know

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    here is what i gathered up from 1st yr notes if this helps consolidate what dan is saying

  26. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  28. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  29. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh because

  30. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    if p is a factor of m^2 then u must have p^2 in m^2

  31. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    other wise its not possible

  32. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    for any square integer there have to be atleast 2 of every prime

  33. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Alright, that's something that needs to be proven or otherwise justified. That's not a field axiom or anything.

  34. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  35. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh im not sure about axioms and stuff

  36. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the only thing i know is the basic principle that all integers are products of primes

  37. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  38. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The relation \[pn^2=m^2\]Would imply to me that m^2 must be divisible by p. Why is this not correct.

  39. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    http://prntscr.com/8cr1bj do u understand this conclusion

  40. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    how that certain statement has to gurantee an existance of some factor * p^2 in it

  41. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    since m and n are integers

  42. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  43. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  44. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Lol. Suddenly random variables.

  45. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    im just saying it has to be some otehr factor of p,

  46. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    let k1^2 and k2^2 be some random factors

  47. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[p=\frac{m^2}{n^2} \ \ \ \text{Agreed.}\]\[pn^2=m^2 \ \ \ \text{Yup.}\] \[\text{m^2 is equal to n^2p, so p must be a factor of m^2.}\]\[m^2=pk.\]

  48. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    tell also k happens to be a perfect square

  49. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    p^2 must be a factor of m^2 actually

  50. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    but yes p too

  51. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    okay lets work from here , this seems to be the most confusing part p*n^2=m^2

  52. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    now lets start with the basic definition of m

  53. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  54. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1441422178020:dw|

  55. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    since m is an integer it can always be written has some product of primes

  56. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  57. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1441422303581:dw|

  58. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  59. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    that is multiplication

  60. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    every integer is a product of primes is what im saying

  61. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    every integer product of primes with some exponent

  62. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    for example 25=5^2 15=3*5 15^2=3^2*5^2

  63. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    notice how every squard integer will have atleast 2 exponent on every prime

  64. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    its always a multiple of 2

  65. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    are u good with this step now?

  66. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    p*n^2=m^2 then p*n^2=p^2*k^2

  67. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  68. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Alright, I think I get your reasoning underlying this specifically, wait a minute.

  69. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  70. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  71. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yeah

  72. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Alright, cool. Agreed.

  73. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    okay now ther is an intermediate step i didnt state cuz its kidn of implied

  74. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  75. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    what we did earliier was very similiar but turns out we could have stopped at this step itself

  76. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  77. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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:

  78. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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"?

  79. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What is the "first statement" exactly, lol.

  80. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    http://prntscr.com/8cr8qz

  81. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  82. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    any questios?

  83. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    theres actually another contradiction argument for completeness sake

  84. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  85. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  86. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  87. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    And thank you so much for your help, you're really good at this.

  88. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    like okay we saw that for ever square all the exonents must be factors of 2

  89. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah, agreed.

  90. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  91. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    so the left side will always have odd* exponent for p

  92. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  93. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh which 2nd one you mean the one in the screenshot?

  94. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    http://prntscr.com/8cr8qz

  95. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah

  96. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    so if \[n^2=p*k_1^2\] then \[n^2=p^2*k_2^2\]

  97. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    just like how \[m^2=p*n^2\\ \] \[m^2=p^2*k^2\]

  98. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    both cases we are showing that n^2 and m^2 had a p^2 in therm which cannot be

  99. Mendicant_Bias
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  100. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.