## Mendicant_Bias one year ago (Introductory Real Analysis) I'm trying to show by contraposition that if the square of a number has a certain number as a factor, its root will also have that certain number as a factor. The way I'm tempted to solve this (after taking the contrapositive of the above statement) is by using a not equal sign, but I'm not sure I understand the mechanics of it. Math below shortly.

1. Mendicant_Bias

Contrapositive of the above: $\text{If n is not a factor of} \ a, \ \text{then n is also not a factor of} \ a^2.$ $a \neq nk;$$a^2 \neq n^2k^2$

2. Mendicant_Bias

I'd like to treat a not equals sign as a relationship that is maintained under normal algebraic operations, e.g. if you manipulate both sides in the same way, the relationship of "not holding"...will still hold. I don't know if I can do this, though.

3. zzr0ck3r

Are we assuming $$a\in \mathbb{Z}$$?

4. Mendicant_Bias

I'm trying to do this specifically by contraposition, but we can do that, too, if we're doing what I think we're doing. And I don't think it should matter for this problem, I think a is an element of either Z or R. I don't totally understand the implications of what you wrote, I'm taking a minute to think about it.

5. zzr0ck3r

Well, to be factor-able...

6. ganeshie8

the term, "factor", tells us that $$a\in \mathbb{Z}$$

7. Mendicant_Bias

Yeah, this is actually just a lemma I'm having to prove in the middle of another problem to complete the whole problem; the original prompt doesn't specify the set the value is in, but the original problem itself is proving that sqrt(3) is irrational; I did it by contradiction, so within the context of a proof by contradiction, I should be assuming that sqrt(something) is rational, if that's sensible/correct?

8. Mendicant_Bias

And yeah, a in this context is supposed to be part of a ratio of two integers in lowest terms, so it definitely belongs to Z. Had to think that through for a minute, sorry, lol.

9. zzr0ck3r

There is an easier way.

10. ganeshie8

$$\mathbb{Z}$$ is the set of integers $$\mathbb{Q}$$ is the set of ratio numbers (rational)

11. ganeshie8

the phrase "factor of a rational number" makes no sense

12. ganeshie8

factor must refer to integer $$a$$ must be an integer, then only we can talk about its factors/divisors

13. Mendicant_Bias

Well, at least for proving the a^2 factor part (I brought it to my prof), I have to do that by contraposition, but I'd ideally like to learn any and all ways to solve this. In any case, I hate to be like, "bluh, this is semantics", but I think we're on the same page regarding what the rationals/integers are, I'm just being really imprecise with my language and throwing stuff together quickly to try and solve this.

14. Mendicant_Bias

Yes, and a is an integer. No disagreement there. But it's both an integer and rational, right?

15. ganeshie8

all integers are rational whats exciting about that haha

16. Mendicant_Bias

And you have to assume both for the problem-right? You have to assume that a and b are integers since it's a ratio of two integers that makes a rational number that you're assumign sqrt(something) is equal to for the sake of contradiction. Nothing, but you said that "factor of a rational number" doesn't make sense, but it does, doesn't it? Integers are rational numbers, and they have factors. Even in lowest terms. It's trivial, but it's not wrong, and that wasn't the point, in any case. People were asking me questions regarding what set the variable we're concerned with was in, and I was answering them. It's not really that important right this very moment, but I was just trying to answer questions I was asked so we could move on.

17. zzr0ck3r

Suppse $$\sqrt{3}=\frac{a}{b}$$ where $$(a,b)=1$$. Then $$3=\frac{a^2}{b^2}\implies b^2*3=a^2$$ from here you may conclude that $$a^2,b^2$$ are both odd. Suppose $$b^2$$ was even, then so is $$3b^2$$ and then so is $$a^2$$ so that we must have $$a,b$$ are both even contradiction $$(a,b)=1$$ So $$a^2,b^2$$ are both odd, so that $$a,b$$ are both odd. So $$a=2k+1\\ b=2k_0+1$$ now plug this in, $$3b^2=a^2\implies 3(4k^2+4k+1)=4k_0^2+4k_0+1\\\implies 6k^2 + 6k + 1 = 2(k_0^2 + k_0).$$ The left is odd, the right is even. Yikes!!!

18. zzr0ck3r

$$(a,b)=1$$ just means that $$\frac{a}{b}$$ CAN'T be reduced any further.

19. Mendicant_Bias

^Part of what I'm not on board with with this/would need to be proved is this: http://i.imgur.com/8kyErQo.png I'm aware that the square of an odd number is odd, and the square of an even number is even, but I don't know how we can conclude that both a^2 and b^2 themselves are odd.

20. Mendicant_Bias

What I'd conclude from 3b^2=a^2 is that 3 is a factor of a^2. I don't yet know, however, how I can demonstrate that both a^2 and b^2 are odd.

21. zzr0ck3r

Ok well you also must conclude that $$a^2$$ has the same parity as $$b^2$$ right?

22. Mendicant_Bias

Wait wait wait, sorry, lol. One minute. I scanned it the first time you posted it, my bad.

23. zzr0ck3r

forget what I just posted, I thought you had another problem.

24. zzr0ck3r

back to your issue. You agree that if $$b^2$$ is even then so is $$a^2$$ right?

25. zzr0ck3r

because $$a^2=3*$$even

26. Mendicant_Bias

Yeah.

27. zzr0ck3r

Ok so let us assume they are both even, then you also agree that if $$a^2$$ is even, then so is $$a$$?

28. Mendicant_Bias

But we don't know that b^2 is even. (Nevermind, assumption) And yeah, agreed that if a^2 is even, so is a.

29. zzr0ck3r

They must either be both even or both odd. I will assume they are both even and show you that that cant be the case. there for they must both be odd. Make sense?

30. Mendicant_Bias

Alright, yeah.

31. zzr0ck3r

OK so if $$a^2,b^2$$ is even, then so is both $$a,b$$ and then re can reduce our fraction $$\frac{a}{b}$$ right?

32. zzr0ck3r

re=we

33. Mendicant_Bias

Yeah.

34. zzr0ck3r

But we assumed at the start that our fraction was reduced so this is a contradiction.

35. zzr0ck3r

So $$a^2,b^2$$ are both odd.

36. zzr0ck3r

p.s. this proof got a dude killed :)

37. zzr0ck3r

Maybe the smiley was too much

38. Mendicant_Bias

Oh, I heard about that! What's his name, the dude who the Pythagoreans killed for disclosing it when they wanted to keep it a secret. And yeah, I get it now, at least I certainly get the proof that a and b must both be odd. Now I'm going to re-read the above, one sec.

39. zzr0ck3r

Well they did not want him to show the world that there were irrational numbers because they thought the world of integers and really did not like that idea. So death ... they also did not allow an open flame next to a mirror...goes to show what they know.

40. Mendicant_Bias

I'm just figuring out the last bit of algebra, but it at the very least conceptually makes sense to me...thanks very much.

41. zzr0ck3r

the algebra will be easy. np.

42. Mendicant_Bias

Yeah, I don't understand what the heck you did at the very end, it literally just looks wrong, so let me write this out step by step and confirm I'm not crazy:

43. Michele_Laino

the starting question is not proved yet :)

44. Michele_Laino

since n can be any natural number, and not n=3

45. zzr0ck3r

lol omg. actually what we proved has nothing to do with the starting question. The starting question is a result of going a different rout that is not needed.

46. Mendicant_Bias

$3(4k^2+4k+1)=4k_0^2+4k_0+1$(I'm on board with this) $12k^2+12k+3=4k_0^2+4k_0+1$$12k^2+12k+2=4k_0^2+4k_0$$6k^2+6k+1=2k_0^2+2k_0$ Alright, so this is effectively the same results-I just don't know...how you did your wizard algebra to end up with a different simplification.

47. zzr0ck3r

I did the exact same thing. I just skip steps...

48. Mendicant_Bias

You had a slightly different written end result, promise. You had plus one on the RHS, which is why I was really confused. Both sides had a plus one, guessing it was a typo. Either way, this makes sense now, heh.

49. Mendicant_Bias

Whoah, hey there, lol. I like hearing everybody here talk. ;_;

50. zzr0ck3r

It was a personal jab at him, just playin.

51. zzr0ck3r

You are right :)

52. Michele_Laino

a good idea would be if you answer to the actual question :) @zzr0ck3r

53. zzr0ck3r

read man, we covered all of this...

54. Mendicant_Bias

ooooooOooo shots fired

55. zzr0ck3r

Or tell us again how $$n$$ might not be $$3$$....

56. Michele_Laino

I tell you that chemistry is not physics, as you wrote yesterday, lol! :) @zzr0ck3r

57. zzr0ck3r

This dude follows me to other questions and trys to but in and ALWAYS says the wrong thing. Then I show him why he was wrong, as I did here, he then moves on to something else. What you are referring to is someone asking a chem question in the math section and saying the reason was because people would not answer it is physics. I told the user that people were not answering because his question was chem, not physics. Just read @Michele_Laino , you might then actually know what you are talking about more.

58. Mendicant_Bias

So at least for the original question, for n=3, we're good, right? We proved all the lemmas necessary in between about how both a and b must be odd, and provided a contradiction showing that sqrt(3) is irrational, but how could we show that if a^2 has a factor alpha, then a also has a factor alpha. The last thing I'm concerned with (and I'll open up a different question if people would like) is just figuring that out.

59. zzr0ck3r

Ahh I did not read that post correcty. I did not know the number was a square... sorry. this is true then.

60. ganeshie8

I like one line proofs in number theory below is a legitimate proof for showing$$\sqrt{3} \not\in \mathbb{Z}$$ : suppose $$\sqrt{3}$$ is rational, then for some $$a,b\in \mathbb{Z}$$ we have $\frac{a}{b} = \sqrt{~3~} \implies \left(\frac{a}{b}\right)^2 = 3 \implies 3b^2 = a^2$ By euclid lemma, $$(a,b)=1 \implies 3\mid 1$$, contradiction $$\blacksquare$$

61. zzr0ck3r

much nicer

62. ganeshie8

I think we still need to show $$(a^2, b^2)=1 \iff (a,b)=1$$

63. ganeshie8

** I like one line proofs in number theory below is a legitimate proof for showing$$\sqrt{3} \not\in \color{red}{\mathbb{Q}}$$ : suppose $$\sqrt{3}$$ is rational, then for some $$a,b\in \mathbb{Z}$$ we have $\frac{a}{b} = \sqrt{~3~} \implies \left(\frac{a}{b}\right)^2 = 3 \implies 3b^2 = a^2$ By euclid lemma, $$(a,b)=1 \implies 3\mid 1$$, contradiction $$\blacksquare$$

64. Mendicant_Bias

Wait a minute, I don't understand, what did we still need to show? Up to the point I thought I was done, what else did we need to prove?

65. ganeshie8

we're done!im talking about my proof in the end... it has nothing to do with zzr his proof is complete

66. zzr0ck3r

you are done, this is a different method. But you need euclids lemma for this version

67. zzr0ck3r

Number theorist like to make things easy.... pfft

68. zzr0ck3r

jk :) this one is much "nicer"

69. Mendicant_Bias

Alright. The last thing I'd like to double down on again is that I want to prove that a factor of a^2 is also a factor of a, and I'd like to do that doing contraposition, but I don't know how to go about doing that. Solving one problem in as many ways possible is right now pretty important to me so that I develop a sense of the different ways of solving problems, not just getting the solution.

70. zzr0ck3r

The best proofs are one liners. Here is my favorite proof that the identity element is unique in a group. $$e_1=e_1*e_2=e_2$$

71. zzr0ck3r

Can I see how you are using this in the proof that $$\sqrt{3}$$ is irrational?

72. Mendicant_Bias

(Hell, that *was* my original question. I was focusing on that lemma which ended up never getting answered, but the larger question getting answered in a different way.) @zzr0ck3r , sure. One minute.

73. Mendicant_Bias

How do you guys put (not LaTeX text) plaintext in line with math in openstudy comments, by the way?

74. zzr0ck3r

$$\text{like this}$$

75. ganeshie8

Actually I am using that in my proof above while making the jump : $$(a^2,b^2)=1 \implies (a,b)=1$$

76. zzr0ck3r

This is some number theory stuff, @ganeshie8 will love to do it :)

77. zzr0ck3r

Ah, I would go a different rout for that.

78. zzr0ck3r

if some number divides a and b then surely it divides a^2 and b^2...

79. Mendicant_Bias

1.) Assume sqrt(3) is rational (proof by contradiction.) $\text{If} \ \sqrt{3} \ \text {is rational, then} \ \sqrt{3}=\frac{a}{b}, \ \ \ a,b,\in \mathbb{Z}$$3=\frac{a^2}{b^2}$$3b^2=a^2$ @zzr0ck3r Yeah, but I want that to be proven deductively, lol, not just going "lol it's obvious". If a^2=3b^2, 3 is a factor of a^2. If b is already assumed as an integer, its square will also be an integer (here's where this lemma is necessary).$3b^2=a^2 \rightarrow 3b^2=(3 \gamma)^2=9 \gamma^2$

80. Mendicant_Bias

$3b^2=9 \gamma^2$By the same logic, 3 is a factor of b^2. By the same lemma as before, if 3 is a factor of b^2, it must be a factor of b. Thus, b=3(delta), where delta is some integer. Putting these into the original expression a/b shows you that they were not, as assumed, in lowest terms. This is the contradiction that shows you that sqrt(3) is not rational.

81. Mendicant_Bias

And Ganeshie, I actually have no idea what that notation means in your last post/how would that be read in words?

82. ganeshie8

$$(a, b)$$ is read as $$\gcd$$ of $$a$$ and $$b$$

83. Mendicant_Bias

But yeah, that's why I'm trying to prove-by contraposition-that if a^2 has some root, a must also have it in common. Ah.

84. Michele_Laino

first: I don't follow you, since I have no reason second: what I write is correct, furthermore, what I saw is that you are very adept at distorting the truth and in offending others, and less capable in doing mathematics, it is clear that your parents have not been able to teach good manners. @zzr0ck3r

85. Mendicant_Bias

Guise pls.

86. ganeshie8

i smell the mathematician vs physicist fight

87. Michele_Laino

@ganesh he started to offend me as you can see from the above replies

88. Michele_Laino

I promised to Mrs. Preetha that I would not fight with that person, but he continues to offend me

89. Mendicant_Bias

...I just want to get my original question answered. Could this please be done somewhere else. I still haven't figured out the question I first asked and was first and foremost was concerned with, everyone sort of ignored that, went, "No, I have a better method" and did whatever they felt like. Which is cool, but I still don't understand how to do this in this specific way that I'm concerned with.

90. Michele_Laino

right! In fact I wrote that the starting question is not proved yet @Mendicant_Bias and @zzr0ck3r started to offend me

91. ganeshie8

 if the square of a number has a certain number as a factor, its root will also have that certain number as a factor. is equivalent to showing : if the $$\gcd(a^2,b^2)$$ is $$d^2$$, then the $$\gcd(a,b)$$ is $$d$$

92. ganeshie8

lets try proving the latter statement..

93. Mendicant_Bias

...What? Alright, hold off, I didn't try to offend you at all. I didn't say a word to you. I don't have an issue with you.

94. Mendicant_Bias

Don't bring me into this. Anyways, yeah, let's try proving that.

95. zzr0ck3r

@Michele_Laino again, simply read " but I'd ideally like to learn any and all ways to solve this." -TheQuestionAsker just read man, read.

96. Mendicant_Bias

Alright, so I have no idea how to compare those two statements off the top of my head, hmm...

97. ganeshie8

do you have teamviewer/skype ?

98. Mendicant_Bias

And well wait a minute, I know it's in effect the same, but I don't need both a^2 and b^2, just doing the same thing for one number in the same set should apply to both right? So If$\text{If} \ \gcd(a^2,b^2)=d^2, \\ \text{then}\gcd(a,b)=d$Is effectively the same as$\text{If} \ \gcd(a^2)=d^2, \ \text{then} \ \gcd(a)=d.$

99. Mendicant_Bias

I do have skype where I am, but I wouldn't want to talk over the mic-it's pretty late where I am and my roommates are sleeping. In any case, I might call it a night unless I can figure this out soon, I'm getting pretty tired. Thank you guys for all the help in any case.

100. ganeshie8

gcd = greatest "common" dividsor gcd requires at least two operands

101. Mendicant_Bias

Operands?

102. Mendicant_Bias

103. Mendicant_Bias

Yeah, I'm too sleepy to think well, I'm going to call it a night and come back to this in the morning. Goodnight and thanks.

104. ganeshie8

gnite! have good sleep :)

105. mathmate

I am sorry that I did not follow the whole thread in detail, but I wish I was there at the beginning. So please point it out if I am repeating some ideas presented before. As @michele-laino stated, the original statement is not proved yet. In fact, it is false, namely "that if the square of a number has a certain number as a factor, its root will also have that certain number as a factor." if a|b^2k^2 then a|bk is false. (counter example: if "16|144, then 16|12 " ) So it's contrapositive is also false. As @ganeshie8 said, we should only be dealing with Z, for example: Perhaps the original statement meant to be. "that if the square of a number $$n\in Z$$ has a certain $$prime~k \in Z$$as a factor, the (root of n) $$\in Z$$ will also have k as a factor."

106. Michele_Laino

thanks! :) @mathmate