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ParthKohli

Please explain me the Euclidean Algorithm. Help appreciated.

  • one year ago
  • one year ago

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  1. KingGeorge
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    First step, look at an example. I'll color things to make it easy to see what I'm doing.

    • one year ago
  2. ParthKohli
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    Okay - I follow.

    • one year ago
  3. KingGeorge
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    Look at \(a=572\) and \(b=165\). The euclidean algorithm will give us the gcd of these two numbers. Here's how you would find it \[572=\color{red}{165}\cdot3+\color{green}{77}\]\[\color{red}{165}=\color{green}{77}\cdot2+\color{blue}{11}\]\[\color{green}{77}=\color{blue}{11}\cdot7+0\]Now that that last number is 0, look at the remainder above it. In this case, that's 11. Hence, \(\gcd(572,165)=11.\)

    • one year ago
  4. KingGeorge
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    The last number in each of those lines is called the "remainder" and the black numbers (3,2,7) are the quotients. In general, given any two numbers \(a,b\), they can be written as \[a=b\cdot q +r\]where q=quotient and r=remainder.

    • one year ago
  5. ParthKohli
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    So the point is that we must repeat it again and again till we get the remainder 0? That does make sense.

    • one year ago
  6. ParthKohli
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    May I have a practice problem?

    • one year ago
  7. KingGeorge
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    Try it out on \(a=342\) and \(b=295\).

    • one year ago
  8. KingGeorge
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    One more thing, make sure \(0\leq r<b\) in the equation \(a=b\cdot q+r\).

    • one year ago
  9. asnaseer
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    @KingGeorge - wonderful explanation! :)

    • one year ago
  10. ParthKohli
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    I wish that the Chrome Aw Snap didn't exist.

    • one year ago
  11. Ishaan94
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    OMG I never knew it's called euclidean algorithm. I love this method. lol

    • one year ago
  12. ParthKohli
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    \[342= 295\cdot 1 + 47 \]\[295 = 47 \cdot 6+13 \]\[47 = 13\cdot 3+8 \]\[13 = 8\cdot 1 + 5 \]\[8 = 5 \cdot 1 + 3 \]\[5 = 3 \cdot 1 + 2 \]\[ 3 = 2 \cdot 1 + 1\]\[2 = 1 \cdot 2 + 0 \]

    • one year ago
  13. ParthKohli
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    So, they are co-prime!

    • one year ago
  14. ParthKohli
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    Thank you :)

    • one year ago
  15. ParthKohli
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    Let me return back to that CRT question. Okay.

    • one year ago
  16. KingGeorge
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    You're welcome. If you want a proof of the algorithm, I could probably type that up as well.

    • one year ago
  17. ParthKohli
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    Haha no :P

    • one year ago
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