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UnkleRhaukus

  • 2 years ago

\[(\forall n\in \mathbb Z)[n|n^2]\]

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  1. UnkleRhaukus
    • 2 years ago
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    t/f

  2. ParthKohli
    • 2 years ago
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    What does "|" mean in logic?

  3. joemath314159
    • 2 years ago
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    divides

  4. ParthKohli
    • 2 years ago
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    The division thing? Yes.

  5. ParthKohli
    • 2 years ago
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    Except for 0.

  6. swissgirl
    • 2 years ago
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    True

  7. ParthKohli
    • 2 years ago
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    What if \(\rm n = 0\)?

  8. swissgirl
    • 2 years ago
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    n^2=n*n so (n*n)/n=n

  9. ParthKohli
    • 2 years ago
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    False.

  10. swissgirl
    • 2 years ago
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    Truesomeness

  11. ParthKohli
    • 2 years ago
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    Though the statement is true for \(\mathbb{Z}^+\)

  12. joemath314159
    • 2 years ago
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    Dont read :\[n\mid n^2\]as division. By definition, \[a\mid b \iff b=ak ,k\in \mathbb{Z}\]there is no division taking place. By this definition, 0 does divide 0^2 since 0^2=0*0

  13. ParthKohli
    • 2 years ago
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    I've always heard, zero cannot divide anything.

  14. joemath314159
    • 2 years ago
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    You are thinking of fractions. Notice the definition doesnt contain anything about fractions.

  15. joemath314159
    • 2 years ago
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    its a statement only about multiplication.

  16. ParthKohli
    • 2 years ago
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    Oh.

  17. UnkleRhaukus
    • 2 years ago
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    \[a|b \iff \exists q[b=aq],a\neq 0\]

  18. joemath314159
    • 2 years ago
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    ah, then i am mistaken. if a cant be zero then what i posted is wrong.

  19. ParthKohli
    • 2 years ago
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    So, in fact, we do have fractions in the definition.

  20. joemath314159
    • 2 years ago
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    hmmm, i still wouldnt say there are fractions. This is generally the way they talk about division in Rings, where only multiplication and addition are defined. But yes, a cannot be zero. http://primes.utm.edu/glossary/xpage/divides.html

  21. UnkleRhaukus
    • 2 years ago
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    \(a|b\) denotes \(b\) is divisible by \( a \)

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