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  • one year ago

a, b, and c are special numbers...

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  1. anonymous
    • one year ago
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    Yeah thats what my mom says

  2. Empty
    • one year ago
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    lol They're positive integers, and when you raise them to positive integer exponents there's only one possible way to get that number: \[\Large a^xb^yc^z\] So for instance, a=4, b=6, and c=9 wouldn't work since \[4^1*6^0*9^1 = 4^0*6^2*9^0\] So this isn't a unique representation! However this will work as a possible value: a=12, b=18, and c=5 So what's the smallest interval that I can always pick all the values of a, b, and c from?

  3. anonymous
    • one year ago
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    for tomorrow!

  4. anonymous
    • one year ago
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    or next week xD

  5. ganeshie8
    • one year ago
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    .

  6. dehelloo
    • one year ago
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    -have to do presentation for class -start working out -zyzz is my inspiration -presentation again -trembling -just keep telling myself "i'm fawkin zeez bruh" -get confident -my turn -i get up there -start shaking uncontrollably -start telling myself "i'm fawkin zeez bruh" -teacher says I can start anytime -I start off with "i'm fawkin zeez bruh" -at this point I'm so nervous I blackout -"i'm fawkin zeez bruh" -repeat at least 4 more times -look around the room, people are saying "why does he keep saying that?" -girls start laughing -I pass out -hit head on the corner of teacher's desk -minor concussion -teacher thinks I was on drugs -classmates call my zeezprah -nickname eventually turns into zebra -i haven't heard my real name in months -haven't been this depressed since high school

  7. Loser66
    • one year ago
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    Question: positive integers includes 0?

  8. Loser66
    • one year ago
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    x, y, z can be the same, but if they are not 0, and a, b, c are distinct, I think it is only one way to get the form a^x b^y c ^z

  9. Empty
    • one year ago
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    x,y,z can be 0, 1,2, etc... but a, b,c have to be 2,3,4,5... etc only

  10. Loser66
    • one year ago
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    How about primes?

  11. Loser66
    • one year ago
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    If they are primes, although x, y , z can be the same, \(a^x\neq b^x\neq c^x\) Hence we never get the same numbers.

  12. ytrewqmiswi
    • one year ago
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    yea its like if u prime factorize any number lets say N then u get -\[N=a^xb^yc^z\] where a,b,c are primes and thats the only way to get that number.. but ur talking about an interval... m confused with the interval thingy.. will it be the smallest interval in which 3 prime numbers lie??..

  13. Empty
    • one year ago
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    Yeah that's an excellent observation. @ytrewqmiswi and it's definitely related. However primes aren't the only way you can do this, which is why the interval is not necessarily primes.

  14. ytrewqmiswi
    • one year ago
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    so we want an interval from which if we pick any three numbers a,b,c then and then on their multiplication we get a number which can be obtained only in that way..?

  15. Empty
    • one year ago
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    Yeah, this is hard to wrap my mind around too haha. So let's just say that the interval is 10. If I give you the number 151 then you can pick two numbers between 151 and 161 so that it satisfies this. But these are specific results the interval could be some other number or not exist at all even.

  16. Empty
    • one year ago
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    For instance, if I only asked for: \[\large a^xb^y\] The answer would be fairly trivial with an interval of 2. Why? a and a+1 are relatively prime, so no chance of them ever having alternate ways to raise them to exponents: \[\large a^x(a+1)^y\]

  17. ytrewqmiswi
    • one year ago
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    oh ok got that let me think some more :)

  18. ytrewqmiswi
    • one year ago
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    is the answer is 4? like we have 2,3,4,5 ...

  19. ytrewqmiswi
    • one year ago
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    well i jst said that the interval is [2,5] cause its the smallest one.. do we need an interval like that in the case where we took 2 numbers and then where ever u go the interval is 2 where ever u go refers to ---> u go to the number 73 and stand on it nd then say that the interval is 2 and if u go to anyother number lets say 37 then also the interval is 2 so do we need a minimum interval which we put anywhere and it works ?

  20. Empty
    • one year ago
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    Yeah precisely, minimum interval we can put anywhere and it works. tricky! You're allowed to look at larger numbers since small numbers like 2,3,4,5, etc... are actually kind of like special cases that you can avoid.

  21. ytrewqmiswi
    • one year ago
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    yes :) ok now i got it :D

  22. ytrewqmiswi
    • one year ago
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    wait we know that the minimum value of the interval can be 4 and 4 is obviously not the answer.... so lets say 7 is the answer and we stand on 10 we got the numbers [10,16] we take any three numbers a,b,c as 10, 12, 14 and we take x,y,z as 3,2,1 so \[a^xb^yc^z=10^312^214^1 =(2\times5)^3(2^2\times3)^2(7\times2)=2^83^25^37\]so basically (a^x)(b^y)(c^z) is not a unique representation ....so no such interval exists that could be put anywhere ...and also there can't be any such unique representation until and unless the numbers are prime

  23. ytrewqmiswi
    • one year ago
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    and also if we consider a case where we had only a and b lets say we stand on 3 we have [3,4] if we consider that the interval is 2 and lets take x,y as 3,7 still we can represent it as (2^14)(3^3) so thats not a unique representation

  24. Empty
    • one year ago
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    Oh I see, I think you are interpretting my question wrong since I wasn't precise enough and didn't foresee this, although this is quite interesting on its own! Also there are some legitimate problems I see as well here that don't seem to work out, so I'll try to explain cause this is making my head spin a little. First: I don't think 4 is the minimum interval, but it could be due to us thinking of two different problems so read further on. Second: I'll try to rephrase my question in a new way in terms of 'unique factorization'. Clearly we can have unique factorization if a,b,c are primes, so I'll sorta take this to be what I mean. So for example, \(a=2\), \(b=3\) and \(c=10\) will also give us a unique factorization. If you give me some values of \(x_1,y_1,z_1\) there is no other possible combination of \(x_2,y_2,z_2\) which will give us the same number if these aren't equal. \[\Large a^{x_1}b^{y_1}c^{z_1} \ne a^{x_2}b^{y_2}c^{z_2}\] The main thing is we are considering this 'basis' on its own terms and not looking at other possible values a, b, and c can be. So I think I saw you comparing two separate choices of a,b, and c together, but really I am only concerned with exponent representations _given_ a choice of a,b,c. So hopefully this clarifies that! So to make more clear, even though a=2 and c=10, there's no problem here. However suppose we had a=10 and c=100 then we would have an issue. This is one of several ways we can have a problem. Another way would be to have some other way of combining multiples together, for example this is NOT going to work: a=2, b=3, c=6 since the exponents (1,1,0) will be the same as (0,0,1), which isn't unique in this basis. I haven't really spoken about intervals very much because intervals aren't important once you're in one of these basis sets I guess you could call them. I guess this is kind of getting complicated, and I think it might be easier to start thinking in terms of linear algebra since I am really thinking in terms of determinants and logarithms. But that might make my ideas less clear if you're unfamiliar with them!

  25. ytrewqmiswi
    • one year ago
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    ok got it :)

  26. Empty
    • one year ago
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    Ok so I guess maybe I have dragged this out too much or you're still having fun I don't know! :D I'll type up my answer with reasoning though so if you want to see it just ask. The thing is, this is actually a very specific case of something much broader so it doesn't end here. Unfortunately I don't have any answers of how to extend this to the larger cases I really want to consider.

  27. ytrewqmiswi
    • one year ago
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    ok :)

  28. Empty
    • one year ago
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    Alright! n and n+1 are relatively prime! That's to say \[\gcd(n,n+1)=1\] They share no factors, which is what we need in order to have unique factorization. But this is not where our interval is. Let's instead multiply these by 2: 2n and 2n+2 are now our two new numbers which I'll call a and c. 2n+1 which lies in between the two is what I'll call b! So just like earlier, a difference of 1 means they're relatively prime, \[\gcd(2n,2n+1)=\gcd(a,b)=1\]\[\gcd(2n+1,2n+2)=\gcd(b,c)=1\] This is great news for us because this means that no power of b with a or c will give us any problems, they'll be completely independent! However we need to go back to a and c. Now if \(\gcd(a,c)=1\) we would be fine, then we would know without a doubt that none of these numbers share factors, but in fact they do, using the original gcd from earlier, we know: \[\gcd(a,c)=2\] So really, they're mostly relatively prime! So they only share a single 2 in common but otherwise are completely free. So to make this more concrete, suppose we look at just a and c for specific values: \[10^x12^y\] We see that there is only a unique number for each x and y, due to the fact that we can separate it out into the relatively prime parts: \[2^{x+2y}5^x3^y\]. So here we go, the interval is 3. The only time this fails is when you choose n=1 since going back to the original case, n and n+1 being relatively prime is actually sort of too simple!

  29. Empty
    • one year ago
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    The point is to ultimately extend the concepts of unique factorization on primes to extremely large numbers so that we can find large primes easily since they will have a 'relative factorization' as prime as well.

  30. imqwerty
    • one year ago
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    thanks :) for solution

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