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anonymous

  • one year ago

given x ≡ 1 (mod 3). Is x ≡ 4 (mod 3) true?

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  1. ganeshie8
    • one year ago
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    Below is not required but it helps in relating congruences and divisibility \[x\equiv 1 \pmod 3 \implies 3 \mid (x-1) \implies 3\mid [(x-1) - 3] \\\implies 3\mid (x-4)\implies x\equiv 4\pmod{3}\]

  2. UsukiDoll
    • one year ago
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    modulo 3 looks like [0,1,2] though?

  3. UsukiDoll
    • one year ago
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    like on x ≡ 1 (mod 3). it's x is congruent to the remainder 1 in modulo 3. I can see that part... it's x ≡ 4 (mod 3) that's giving me a hard time... because that's x is congruent to the remainder 4 in modulo 3.

  4. ganeshie8
    • one year ago
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    Notice that \(3\mid (3k)\). This is same as saying \(3k\equiv 0 \pmod{3} \). Therefore \(x\equiv x+0\equiv x+3k\pmod{3}\)

  5. anonymous
    • one year ago
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    so basically I can add multiple of 3 on either sides, and it still holds?

  6. anonymous
    • one year ago
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    like x + 3 ≡ 1 (mod 3) holds?

  7. ganeshie8
    • one year ago
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    Yes, congruence is just a neat notation that hides the messy divisibility rules

  8. ganeshie8
    • one year ago
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    \(x\equiv y\pmod{n}\) and \(y\equiv z\pmod{n}\) implies \(x\equiv z\pmod{n}\) transitive property holds, so you can replace \(3\) by \(0\) in modulo 3

  9. Empty
    • one year ago
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    Whenever I use mods I end up just subtracting off the number until I have some number less than what I want, like this: \[9 \equiv 5 \equiv 1 \mod 4 \] So here I just kept subtracting 4 off, rather than trying to think "ok what's the remainder if I divide by 4?" There are a bunch of cute tricks you can use too, like: \[100^{2523672} \mod 3\] just rewrite 100 as 3*33+1 and we have: \[(3*33+1)^{klfaljag} \mod 3\] \[(0*33+1)^{lsdjf0aug} \mod 3\] \[1 \mod 3\]

  10. anonymous
    • one year ago
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    ok. I know it seems kinda trivial but I just want to make sure I really understand it. x + 3k ≡ 1 + 3s (mod 3), for some integer k,s, holds ? right? ^^

  11. anonymous
    • one year ago
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    actually for *all* integer s,k

  12. ganeshie8
    • one year ago
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    x + 3k ≡ 1 + 3s (mod 3) thats not so important, all you need to do is to refer back to the definition : \[a\equiv b\pmod{n}\iff n\mid (a-b)\]

  13. anonymous
    • one year ago
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    @Empty i'm not quite there yet but I might find that useful one day ^^

  14. ganeshie8
    • one year ago
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    the moment you see 3k, and if you're in modulo 3, you see that 3k is divisible by 3 : \[3\mid (3k) \iff 3k\equiv 0 \pmod{3}\] so you replace \(3k\) by \(0\)

  15. Empty
    • one year ago
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    A way to help think of it is all even and all odd numbers can be represented as 2n or 2n+1. So you can see in mod 2, all even numbers are 0 and all odd numbers are 1.

  16. anonymous
    • one year ago
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    @ganeshie8 ah i see. x ≡ 4 (mod 3) x ≡ 1 + 3 (mod 3) and since 3 is a multiple of 3, replace 3 with 0. x - 99 ≡ 1 (mod 3). And since 99 is also a multiple of 3, replace it with 0. and x + 3k ≡ 1 + 3s (mod 3), replace 3k and 3s with 0.

  17. ganeshie8
    • one year ago
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    Yes, but careful, congruence relations don't work in both directions always

  18. ganeshie8
    • one year ago
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    ``` given x ≡ 1 (mod 3). Is x ≡ 4 (mod 3) true? ``` you should start with whats given, not what you want to prove

  19. anonymous
    • one year ago
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    ah yes. I just need to see what you can validly add to either side (or both). So in general, a ≡ b (mod c) implies a ≡ b + kc (mod c) implies a + kc ≡ b (mod c) implies a + kc ≡ b + sc (mod c)

  20. ganeshie8
    • one year ago
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    Yep, adding/subtracting a multiple of "c" to a side(s) wont change the congruence because : \[c\mid (a-b) \implies c\mid [(a-b) + kc]\]

  21. IrishBoy123
    • one year ago
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    "."

  22. anonymous
    • one year ago
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    yep! I understand everything now. Thank you!

  23. ganeshie8
    • one year ago
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    np

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