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  • one year ago

Let's learn and try to prove the ABC conjecture! (No prior knowledge of Inter-Universial Teichmuller theory necessary!)

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  1. Empty
    • one year ago
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    Ok what the hell is the ABC conjecture? First we have to define a function, \[rad(n)\] All it does is gives you a number with the exponents on its prime factorization thrown away leaving only 1s left, like this: \[rad(12)=rad(2^23^1)=2^13^1=6\] Now we're almost ready to state the conjecture. We start with two numbers \(a\) and \(b\), \[a+b=c\] and here's the conjecture! \[c> rad(abc)^k\] This statement is true for only finitely many numbers for any choice of \(k>1\)

  2. Empty
    • one year ago
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    @FireKat97 Ok I'm not sure I fully understand this I guess I'm going to try throwing in random examples to play with it to try to figure it out. :P

  3. Empty
    • one year ago
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    Ok something interesting is if a and b don't share factors, then neither does c, so they're all relatively prime: \[\gcd(a,b)=\gcd(a,c)=\gcd(b,c)=1\] Since that's true I think we can separate out the rad function this way since it won't matter: \[rad(abc)=rad(a)*rad(b)*rad(c)\]

  4. Empty
    • one year ago
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    Anybody interested in trying to figure this out or have any ideas of interesting stuff? I don't wanna feel like I'm yelling in an echo chamber lol.

  5. FireKat97
    • one year ago
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    does it help if I say 'everything above makes sense'? xD

  6. dan815
    • one year ago
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    what do u mean its only true for finitely many numbers for some choice k > 1

  7. dan815
    • one year ago
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    arent there an infinite nuumber for c, that we can pick

  8. Kainui
    • one year ago
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    Prove it dan, then you'll have found a counter example to the abc conjecture lol

  9. Kainui
    • one year ago
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    https://en.wikipedia.org/wiki/Abc_conjecture#Formulations

  10. dan815
    • one year ago
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    like suppose i pick c^n+1, and a=1,b=1, then rad(abc)= c and c^n+1 = c^k now i can pick n=k and get an infinite number right

  11. Kainui
    • one year ago
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    No because a+b=c

  12. dan815
    • one year ago
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    ohhhhhhhhhhh

  13. dan815
    • one year ago
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    ok now i see why there could be finite okay!

  14. Kainui
    • one year ago
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    Haha yeah I feel like this is related to the arithmetic derivative, also did you notice the rad(n) function is multiplicative? Another fun fact although probably useless is this identity you could call it with the mobius function. \[\mu(rad(n)) \ne 0\]

  15. dan815
    • one year ago
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    true

  16. dan815
    • one year ago
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    lol

  17. Kainui
    • one year ago
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    I was looking at this \[c=p^{k+1}\] so that we have \[p > rad^k(ab)\]

  18. dan815
    • one year ago
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    oo ok ok

  19. dan815
    • one year ago
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    a + b = c and a*b <c

  20. Kainui
    • one year ago
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    I guess this probably always false maybe I should have just written this out: \[c=p^k\] \[p^k > rad^k(abp^k)=rad^k(ab) p^k\] \[1 > rad(ab)\] So like because of stuff like this there are infinitely many cases where I see that it must be false it's just proving that there are finitely many cases where it's true, kinda weird.

  21. dan815
    • one year ago
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    c=p^(k+1) p > (ab)^k

  22. Kainui
    • one year ago
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    That's only true if a and b are primes

  23. Kainui
    • one year ago
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    \[rad^y(p^x)=p^y\] It's kinda like a prime renamer huh interesting.

  24. dan815
    • one year ago
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    a+b = p^(k+1) p=root_{k+1} (a+b) <--- this being an integer has some big restrictions

  25. Kainui
    • one year ago
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    wait what do you mean by root_{k+1}

  26. dan815
    • one year ago
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    K+1 th root of A+B

  27. Kainui
    • one year ago
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    ``` \[ \sqrt[k+1]{a+b}\] ``` \[ \sqrt[k+1]{a+b}\]

  28. dan815
    • one year ago
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    yea lol

  29. Kainui
    • one year ago
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    Oh it has some big restrictions but it's not a big deal that it does.

  30. Kainui
    • one year ago
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    Just start here: \[p^{k}=\frac{p^k+r}{2} + \frac{p^k-r}{2}\] there are always going to be multiple choices of r that work.

  31. dan815
    • one year ago
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    u there?

  32. Kainui
    • one year ago
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    Yeah I'm here I was watching some youtubes what's up

  33. dan815
    • one year ago
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    why are you doing that expression with r

  34. Kainui
    • one year ago
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    I think I kinda made that weirder than it needed to be, I was just showing we can always find a value of a and b that work, \[c=p^k\]\[a=p^k+r\]\[b=p^k-r\] as long as r isn't divisible by p, then all 3 numbers are relatively prime.

  35. dan815
    • one year ago
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    ohh i see! ok ok

  36. Kainui
    • one year ago
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    cause you were trying to say that there are restrictions or whatever. I guess I was looking for an infinite number of counter examples I think I'm partly confused too haha

  37. dan815
    • one year ago
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    /2

  38. Kainui
    • one year ago
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    Yeah ok the divide by 2 has to be in there right I forgot

  39. dan815
    • one year ago
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    okay hmm and this is can even be the maximum one to find

  40. dan815
    • one year ago
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    this should produce the largest or close to largest answers from

  41. dan815
    • one year ago
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    rad(abc) for small rs

  42. Kainui
    • one year ago
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    I was thinking that but I realized it's flawed

  43. Kainui
    • one year ago
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    cause it's not so much the size of the number it's the amount of unique primes in the number that matters more

  44. dan815
    • one year ago
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    oh thats ttrue

  45. Kainui
    • one year ago
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    Also this way isn't really the only way we can pick numbers a and b where they're some distance from the middle, we could pick like \[p^k = (p^k-q^r)+(q^r)\] or something like this idk

  46. Kainui
    • one year ago
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    Ooooh ok I got an idea \[p^k = q \# + (p^k-q \#)\] where q# is the largest primorial not exceeding \(p^k\) this should help us maximize the rad function!

  47. dan815
    • one year ago
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    |dw:1444290020985:dw|

  48. Kainui
    • one year ago
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    Actually one small thing to fix, is if that's true then this definitely means q# is divisible by p... Or maybe it means its likely, so I'll just write: \[\frac{q \#}{p}\]

  49. Kainui
    • one year ago
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    Oh you can't do that cause we require that \(\gcd(a,b)=1\) and here they're both divisible by 2 so that won't work

  50. dan815
    • one year ago
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    aww dang

  51. Kainui
    • one year ago
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    but I like where you're going with that, for a second I was like about to throw my crap down and start working on that cause mixing addition and multiplication like that would be cool. Maybe we can use that idea though somehow.

  52. dan815
    • one year ago
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    ya ppretty much had an answer without gcd condition because uc an jsut do 2^(A+1) > 2^3K from there

  53. dan815
    • one year ago
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    but i guess thats why this inequality has finite because of that exact gcd reason

  54. dan815
    • one year ago
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    how about something like that though lets take smallest possible primes

  55. dan815
    • one year ago
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    2^a + 3^b= 2^N

  56. dan815
    • one year ago
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    does c also have to be gcd 1 or just a and b

  57. Kainui
    • one year ago
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    Ok this looks fun and useful. Also if \(\gcd(a,b)=1\) then this implies \(\gcd(a,c)=\gcd(b,c)=1\) so you're good.

  58. dan815
    • one year ago
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    oh that makes sense

  59. dan815
    • one year ago
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    its easy to see especially if u have p1^a + p2^b, if its the sum of 2 diffrent prime exponents

  60. Kainui
    • one year ago
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    proof, suppose b and c share a factor in common, then you can write b=f*m c=f*n a+b=c a+f*m=f*n a=f*(n-m) donezo

  61. Kainui
    • one year ago
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    Ok I took your thing and generalized it slightly, plug in 2 and 3 if you want: \[p^a+q^b=c\] This will give us \[c > rad^k (p^aq^bc)\] \[c > p^kq^k rad^k(c)\] Maybe moving the c's to the same side is nice: \[\frac{c}{rad^k(c)} > p^kq^k\] Or taking the root: \[\frac{\sqrt[k]{c}}{rad(c)} > pq\] Just things to look at that are kinda useful maybe.

  62. dan815
    • one year ago
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    is this ever possible

  63. dan815
    • one year ago
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    p1^a+p2^b=p3^c where p1 p2 and p3 are all primes

  64. dan815
    • one year ago
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    2+3=5 is 1, where abc =1

  65. Kainui
    • one year ago
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    \[3^2+2^4=5^2\]

  66. dan815
    • one year ago
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    ok i like this

  67. Kainui
    • one year ago
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    yeah I was also thinking like maybe we can find some sorta weird contradiction if we look at like \[x+y+z=c\] and then we group them into 3 different ways to form a and b and compare them.

  68. dan815
    • one year ago
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    do abc have to be positive

  69. Kainui
    • one year ago
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    Yeah abc have to be positive integers but k just has to be a real number greater than 1

  70. dan815
    • one year ago
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    hmmmmmmmmmm

  71. dan815
    • one year ago
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    suppose we find a bound

  72. dan815
    • one year ago
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    like c>rac(abc)^1 and c<=rad(abc)^2 then we are done right

  73. dan815
    • one year ago
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    well i guess just finding any possible one is good enough then xD

  74. dan815
    • one year ago
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    ok wait i guess proving there are an inifite right hand sides is not what they want dang it

  75. Kainui
    • one year ago
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    In an attempt to find a contradiction: since \(q\#\) maximizes the rad function, \[max = rad(q \#)\] I pick a+b=c (dividing out p from the primorial so I don't have factors in common) \[ \frac{q\#}{p} + (p^n-\frac{q\#}{p})=p^n\] We get: \[p^n > rad^k(\frac{q\#}{p}(p^n-\frac{q\#}{p})p^n)\] \[p^{n-k} > rad^k(\frac{q\#}{p}) rad^k(p^n-\frac{q\#}{p})\] So to find a contradiction we can show that this is true for an infinite number of cases?

  76. dan815
    • one year ago
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    i think procing it is easier then

  77. Kainui
    • one year ago
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    Yeah actually I keep getting confused on what the hell it is we're trying to hsow

  78. dan815
    • one year ago
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    we know that rad(abc) will have a base > rad(c ) so at some point rad(abc)^k > c, for some k ,after k is a certain value this will always be true

  79. dan815
    • one year ago
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    now we gotta show until k k there will be finite solutiond and that is true because we are allowing only integer abcs

  80. dan815
    • one year ago
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    the last statement is like no work and guess, but i think we have to be able to sho that statement

  81. dan815
    • one year ago
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    lol i am back to begining again i see

  82. Kainui
    • one year ago
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    I got it I think i mixed myself up here, this is what I want to do: \[(\frac{p_n \#}{2}-2^x)+2^x = \frac{p_n \#}{2}\] This will maximize and minimize what we want I think haha.

  83. dan815
    • one year ago
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    okay lemme see what is the number sign again

  84. Kainui
    • one year ago
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    Really like all this means is: \[\frac{p_n \#}{2} = 3*5*7*11*...*p_{n-1}*p_n\] Now let's do some abc's

  85. dan815
    • one year ago
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    okk

  86. dan815
    • one year ago
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    how come u wanna maximize rad function

  87. dan815
    • one year ago
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    oh for contradiction okay

  88. dan815
    • one year ago
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    how are we gonna minimize rad function

  89. Kainui
    • one year ago
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    \[\frac{p_n \#}{2}> rad^k((\frac{p_n \#}{2}-2^x)2^x\frac{p_n \#}{2})\] \[\frac{p_n \#}{2}> rad^k(\frac{p_n \#}{2}-2^x)2^k \frac{(p_n \#)^k}{2^k}\] Wait... \[1> 2 (p_n \#)^{k-1} rad^k(\frac{p_n \#}{2}-2^x)\] This is the opposite of what I wanted to do, this is clearly like a bad case to look at rofl

  90. Kainui
    • one year ago
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    Yeah to minimize the rad function we have to use powers of 2 and 3 I think like you were doing earlier haha. In my mind I was trying to maximize c, but the way I did it ended up not working out the way I intended.

  91. dan815
    • one year ago
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    there are only finite solutions for that apprently

  92. dan815
    • one year ago
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    how about using that

  93. dan815
    • one year ago
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    we know that for all other primes where u dont have 2

  94. dan815
    • one year ago
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    theres no way that p1^a+p2^b=p3^c for a,b,c > 1

  95. dan815
    • one year ago
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    since its odd + Odd =/= odd

  96. Kainui
    • one year ago
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    oh true

  97. dan815
    • one year ago
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    now to see how these things change with more primes thrown in to the mix

  98. dan815
    • one year ago
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    we know all odds thrown in nothings gonna come out

  99. dan815
    • one year ago
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    so we know the existanece of 2 is a must

  100. dan815
    • one year ago
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    one has to have a 2 and the other cannot

  101. Kainui
    • one year ago
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    Like if we restrict ourselves to primes? Ok I see

  102. dan815
    • one year ago
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    yeah not even with prime products now

  103. dan815
    • one year ago
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    no* even with

  104. dan815
    • one year ago
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    maybe we can get something like this

  105. dan815
    • one year ago
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    for a , b , c where a b c are some product of primes, if we can say the number of integer solutions is a function of the number of primes u are using or something like that

  106. Kainui
    • one year ago
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    Yeah I think that's the way to go, something like that sounds cool

  107. dan815
    • one year ago
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    it would be kind of interesting to find a bound for like the numver of interger solutions it cannot exceed a b c

  108. dan815
    • one year ago
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    for a given a b c

  109. Kainui
    • one year ago
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    Yeah definitely, I think looking back at this, it kind of makes sense more to me to look at after this as being a useful form: \[\frac{c}{rad^k(c)} > rad^k(a)rad^k(b)\]

  110. dan815
    • one year ago
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    i got midterm in 5 hours T_T

  111. dan815
    • one year ago
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    help me with quantum tomorrow after my classes!

  112. dan815
    • one year ago
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    i gotta actually study for that exam

  113. Kainui
    • one year ago
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    Haha damn I was just about to say the rad^k function is like a projection operator. Check this out: \[\langle a,b,c \rangle \rightarrow \langle k, k, k \rangle\] Alright sounds fun what chapters in kaye should I try to read or whatever I wanna be on top of it so we can go through it confidently and fast.

  114. dan815
    • one year ago
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    oh its stuff u already know the mid term is only up to last weeks stuff, so its the same stuff we were doing for my assignment i just gotta go over that in detail, ill try to go thru the lecture slide problems, and if we are working fast then we can keep going ahead

  115. Kainui
    • one year ago
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    Ok cool I'm gonna go to sleep and dream about the abc since it's like 4 am haha.

  116. dan815
    • one year ago
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    ok night dood

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