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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.

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  1. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Are we assuming \(a\in \mathbb{Z}\)?

  4. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    Well, to be factor-able...

  6. ganeshie8
    • one year ago
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    the term, "factor", tells us that \(a\in \mathbb{Z}\)

  7. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    There is an easier way.

  10. ganeshie8
    • one year ago
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    \(\mathbb{Z}\) is the set of integers \(\mathbb{Q}\) is the set of ratio numbers (rational)

  11. ganeshie8
    • one year ago
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    the phrase "factor of a rational number" makes no sense

  12. ganeshie8
    • one year ago
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    factor must refer to integer \(a\) must be an integer, then only we can talk about its factors/divisors

  13. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    Yes, and a is an integer. No disagreement there. But it's both an integer and rational, right?

  15. ganeshie8
    • one year ago
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    all integers are rational whats exciting about that haha

  16. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \((a,b)=1\) just means that \(\frac{a}{b}\) CAN'T be reduced any further.

  19. Mendicant_Bias
    • one year ago
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    ^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
    • one year ago
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    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
    • one year ago
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    Ok well you also must conclude that \(a^2\) has the same parity as \(b^2\) right?

  22. Mendicant_Bias
    • one year ago
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    Wait wait wait, sorry, lol. One minute. I scanned it the first time you posted it, my bad.

  23. zzr0ck3r
    • one year ago
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    forget what I just posted, I thought you had another problem.

  24. zzr0ck3r
    • one year ago
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    back to your issue. You agree that if \(b^2\) is even then so is \(a^2\) right?

  25. zzr0ck3r
    • one year ago
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    because \(a^2=3*\)even

  26. Mendicant_Bias
    • one year ago
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    Yeah.

  27. zzr0ck3r
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Alright, yeah.

  31. zzr0ck3r
    • one year ago
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    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
    • one year ago
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    re=we

  33. Mendicant_Bias
    • one year ago
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    Yeah.

  34. zzr0ck3r
    • one year ago
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    But we assumed at the start that our fraction was reduced so this is a contradiction.

  35. zzr0ck3r
    • one year ago
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    So \(a^2,b^2\) are both odd.

  36. zzr0ck3r
    • one year ago
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    p.s. this proof got a dude killed :)

  37. zzr0ck3r
    • one year ago
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    Maybe the smiley was too much

  38. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    the algebra will be easy. np.

  42. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    the starting question is not proved yet :)

  44. Michele_Laino
    • one year ago
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    since n can be any natural number, and not n=3

  45. zzr0ck3r
    • one year ago
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    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
    • one year ago
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    \[3(4k^2+4k+1)=4k_0^2+4k_0+1\](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
    • one year ago
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    I did the exact same thing. I just skip steps...

  48. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    Whoah, hey there, lol. I like hearing everybody here talk. ;_;

  50. zzr0ck3r
    • one year ago
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    It was a personal jab at him, just playin.

  51. zzr0ck3r
    • one year ago
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    You are right :)

  52. Michele_Laino
    • one year ago
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    a good idea would be if you answer to the actual question :) @zzr0ck3r

  53. zzr0ck3r
    • one year ago
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    read man, we covered all of this...

  54. Mendicant_Bias
    • one year ago
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    ooooooOooo shots fired

  55. zzr0ck3r
    • one year ago
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    Or tell us again how \(n\) might not be \(3\)....

  56. Michele_Laino
    • one year ago
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    I tell you that chemistry is not physics, as you wrote yesterday, lol! :) @zzr0ck3r

  57. zzr0ck3r
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Ahh I did not read that post correcty. I did not know the number was a square... sorry. this is true then.

  60. ganeshie8
    • one year ago
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    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
    • one year ago
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    much nicer

  62. ganeshie8
    • one year ago
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    I think we still need to show \((a^2, b^2)=1 \iff (a,b)=1 \)

  63. ganeshie8
    • one year ago
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    ** 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
    • one year ago
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    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
    • one year ago
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    we're done!im talking about my proof in the end... it has nothing to do with zzr his proof is complete

  66. zzr0ck3r
    • one year ago
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    you are done, this is a different method. But you need euclids lemma for this version

  67. zzr0ck3r
    • one year ago
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    Number theorist like to make things easy.... pfft

  68. zzr0ck3r
    • one year ago
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    jk :) this one is much "nicer"

  69. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Can I see how you are using this in the proof that \(\sqrt{3}\) is irrational?

  72. Mendicant_Bias
    • one year ago
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    (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
    • one year ago
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    How do you guys put (not LaTeX text) plaintext in line with math in openstudy comments, by the way?

  74. zzr0ck3r
    • one year ago
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    `\(\text{like this}\)`

  75. ganeshie8
    • one year ago
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    Actually I am using that in my proof above while making the jump : \((a^2,b^2)=1 \implies (a,b)=1\)

  76. zzr0ck3r
    • one year ago
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    This is some number theory stuff, @ganeshie8 will love to do it :)

  77. zzr0ck3r
    • one year ago
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    Ah, I would go a different rout for that.

  78. zzr0ck3r
    • one year ago
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    if some number divides a and b then surely it divides a^2 and b^2...

  79. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    \[3b^2=9 \gamma^2\]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
    • one year ago
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    And Ganeshie, I actually have no idea what that notation means in your last post/how would that be read in words?

  82. ganeshie8
    • one year ago
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    \((a, b)\) is read as \(\gcd\) of \(a\) and \(b\)

  83. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Guise pls.

  86. ganeshie8
    • one year ago
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    i smell the mathematician vs physicist fight

  87. Michele_Laino
    • one year ago
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    @ganesh he started to offend me as you can see from the above replies

  88. Michele_Laino
    • one year ago
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    I promised to Mrs. Preetha that I would not fight with that person, but he continues to offend me

  89. Mendicant_Bias
    • one year ago
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    ...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
    • one year ago
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    right! In fact I wrote that the starting question is not proved yet @Mendicant_Bias and @zzr0ck3r started to offend me

  91. ganeshie8
    • one year ago
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    ` 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
    • one year ago
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    lets try proving the latter statement..

  93. Mendicant_Bias
    • one year ago
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    ...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
    • one year ago
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    Don't bring me into this. Anyways, yeah, let's try proving that.

  95. zzr0ck3r
    • one year ago
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    @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
    • one year ago
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    Alright, so I have no idea how to compare those two statements off the top of my head, hmm...

  97. ganeshie8
    • one year ago
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    do you have teamviewer/skype ?

  98. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    gcd = greatest "common" dividsor gcd requires at least two operands

  101. Mendicant_Bias
    • one year ago
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    Operands?

  102. Mendicant_Bias
    • one year ago
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    (Nevermind, googled it)

  103. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    gnite! have good sleep :)

  105. mathmate
    • one year ago
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    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
    • one year ago
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    thanks! :) @mathmate

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