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MIT PRIMES/Art of Problem Solving

CROWDMATH 2024: Generalizations of the Notion of Primes

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Generalizations of the Notion of Primes J
Exercise 0.3
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Thayaden
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#1
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Crazy Idea but $(M,\gcd )$ I think the condtions work

$$\gcd(3,7)=\gcd(7,3)$$and
$$\gcd(3,7)=\gcd(3,101)$$yet
$$7\neq101$$
This post has been edited 1 time. Last edited by Thayaden, Sep 18, 2024, 6:47 PM
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felixgotti
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What is the monoid $M$ in this context? Not for all monoids the notion of gcd makes sense.
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Thayaden
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I see what you're saying I can't generalise but it definitely holds true for the monoid $(\mathbb{Z},\gcd)$
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Exercise 0.3
Thayaden   2
N Nov 7, 2024 by Thayaden
Crazy Idea but $(M,\gcd )$ I think the condtions work

$$\gcd(3,7)=\gcd(7,3)$$and
$$\gcd(3,7)=\gcd(3,101)$$yet
$$7\neq101$$
2 replies
Thayaden
Sep 16, 2024
Thayaden
Nov 7, 2024
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Exercise 0.3
elee310   0
Jul 29, 2024
What I've tried so far:

$(\mathbb{Z}, *) \text{ is a commutative monoid that is not cancellative.}$
$\text{Ex. } 0*4=0*3 \not \rightarrow 4=3$

Where I'm stuck:
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elee310
Jul 29, 2024
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Exercise 0.3, 0.4
hkoo   0
May 21, 2024
What I've tried so far:

<Describe what you have tried so far here. That way, we can do a better job helping you!>

0.3
$Z/6Z-{0}$ is commutative monoid with multiply operation. for example, 3x2=3x4=0 in this group. however 2 is unequalt to 4. so it's not cancellative.

0.4 Prove that, for each $n \in Z with n>=2$, the ring $Z/nZ$ is an integral domain if and only if n is prime.

(1) only if
if the ring $Z/nZ$ is an integral domain, 0 is a unique zero-divior. If n is composite number, for $d \in Z/nZ$ s.t d | n and d>1, its order is n/d for add operation like d+...+d (add n/d times) = d*(n/d)(by distribution law) = 0. but it's contradiction to assumption s.t $Z/nZ$ is an integral domain. So n is prime number.

(2) if
Let n be a prime number. suppose that there exists non-zero elements r in $Z/nZ$ s.t r is zero-divisor. ie., n divides r*s. however, since $1<=r,s<=n-1$, r*s have no prime factor n. so n does not divide r*s. it's contradiction. conclusively, $Z/nZ$ is integral domain.
Where I'm stuck:

<Describe what's confusing you, or what your question is here!>
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hkoo
May 21, 2024
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Testingg
mitosaurio47   0
Jan 22, 2024
Testingg
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mitosaurio47
Jan 22, 2024
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Exercise 0.3
Thayaden   2
N Nov 7, 2024 by Thayaden
Crazy Idea but $(M,\gcd )$ I think the condtions work

$$\gcd(3,7)=\gcd(7,3)$$and
$$\gcd(3,7)=\gcd(3,101)$$yet
$$7\neq101$$
2 replies
Thayaden
Sep 16, 2024
Thayaden
Nov 7, 2024
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