## 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?

1. anonymous

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

2. dan815

i really dont know what to do yet xD

3. dan815

but im thinking contradiction is the easiest too

4. 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.

5. Mendicant_Bias

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

6. 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).

7. 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).

8. dan815

|dw:1441420377401:dw|

9. dan815

i think this works

10. dan815

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

11. Mendicant_Bias

12. Mendicant_Bias

13. 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

14. Mendicant_Bias

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

15. dan815

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

16. dan815

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

17. dan815

18. dan815

think*

19. 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.

20. anonymous

ye nah ur right dan

21. 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

22. dan815

23. Mendicant_Bias

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

24. dan815

okay LETS write this out agin clearly now that we know

25. anonymous

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

26. 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

27. anonymous

28. 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?

29. dan815

oh because

30. dan815

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

31. dan815

other wise its not possible

32. dan815

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

33. Mendicant_Bias

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

34. 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

35. dan815

oh im not sure about axioms and stuff

36. dan815

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

37. 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.

38. 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.

39. dan815

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

40. dan815

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

41. dan815

since m and n are integers

42. 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?

43. 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

44. Mendicant_Bias

Lol. Suddenly random variables.

45. dan815

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

46. dan815

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

47. 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.$

48. dan815

tell also k happens to be a perfect square

49. dan815

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

50. dan815

but yes p too

51. dan815

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

52. dan815

53. 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.

54. dan815

|dw:1441422178020:dw|

55. dan815

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

56. 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

57. dan815

|dw:1441422303581:dw|

58. 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.

59. dan815

that is multiplication

60. dan815

every integer is a product of primes is what im saying

61. dan815

every integer product of primes with some exponent

62. dan815

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

63. dan815

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

64. dan815

its always a multiple of 2

65. dan815

are u good with this step now?

66. dan815

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

67. 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

68. Mendicant_Bias

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

69. 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.

70. 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.

71. dan815

yeah

72. Mendicant_Bias

Alright, cool. Agreed.

73. dan815

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

74. 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

75. dan815

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

76. 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

77. 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:

78. 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"?

79. Mendicant_Bias

What is the "first statement" exactly, lol.

80. dan815
81. 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

82. dan815

any questios?

83. dan815

theres actually another contradiction argument for completeness sake

84. 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

85. 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

86. 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.

87. Mendicant_Bias

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

88. dan815

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

89. Mendicant_Bias

Yeah, agreed.

90. 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

91. dan815

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

92. 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.

93. dan815

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

94. dan815
95. Mendicant_Bias

Yeah

96. dan815

so if $n^2=p*k_1^2$ then $n^2=p^2*k_2^2$

97. dan815

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

98. dan815

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

99. 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.