UnkleRhaukus
  • UnkleRhaukus
\[(\forall n\in \mathbb Z)[n|n^2]\]
Mathematics
chestercat
  • chestercat
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UnkleRhaukus
  • UnkleRhaukus
t/f
ParthKohli
  • ParthKohli
What does "|" mean in logic?
anonymous
  • anonymous
divides

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ParthKohli
  • ParthKohli
The division thing? Yes.
ParthKohli
  • ParthKohli
Except for 0.
swissgirl
  • swissgirl
True
ParthKohli
  • ParthKohli
What if \(\rm n = 0\)?
swissgirl
  • swissgirl
n^2=n*n so (n*n)/n=n
ParthKohli
  • ParthKohli
False.
swissgirl
  • swissgirl
Truesomeness
ParthKohli
  • ParthKohli
Though the statement is true for \(\mathbb{Z}^+\)
anonymous
  • anonymous
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
ParthKohli
  • ParthKohli
I've always heard, zero cannot divide anything.
anonymous
  • anonymous
You are thinking of fractions. Notice the definition doesnt contain anything about fractions.
anonymous
  • anonymous
its a statement only about multiplication.
ParthKohli
  • ParthKohli
Oh.
UnkleRhaukus
  • UnkleRhaukus
\[a|b \iff \exists q[b=aq],a\neq 0\]
anonymous
  • anonymous
ah, then i am mistaken. if a cant be zero then what i posted is wrong.
ParthKohli
  • ParthKohli
So, in fact, we do have fractions in the definition.
anonymous
  • anonymous
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
UnkleRhaukus
  • UnkleRhaukus
\(a|b\) denotes \(b\) is divisible by \( a \)

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