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Remarks:
Greatest common divisor
Definition - Ideal of Z\mathbb{Z}Z
I⊆Z I \subseteq \mathbb{Z}I⊆Zis an ideal ⟺ ∀ a,b∈I and,z ∈Z\iff \forall \ a, b \in I \text{ and} , z\ \in \mathbb{Z}⟺∀ a,b∈I and,z ∈Zwe have
a+b∈I and az∈Ia + b \in I \text{ and } az \in Ia+b∈I and az∈I
Example: aZ={az : z∈Z}→2Z,3Z,4Z,…a\mathbb{Z} = \{az \ : \ z \in \mathbb{Z} \} \to 2\mathbb{Z}, 3\mathbb{Z}, 4\mathbb{Z}, \dotsaZ={az : z∈Z}→2Z,3Z,4Z,… - multiples of aaa
∀a,b∈Z\forall a, b \in \mathbb{Z}∀a,b∈Zwe have b∈aZ ⟺ a∣bb \in a\mathbb{Z} \iff a | bb∈aZ⟺a∣b
I1+I2={a1+a2 : a1∈I1,a2∈I2}I_1 + I_2 = \{a_1 + a_2 \ : \ a_1 \in I_1 , a_2 \in I_2\}I1+I2={a1+a2 : a1∈I1,a2∈I2} is an ideal
Example: Consider 18Z+12Z18\mathbb{Z} + 12\mathbb{Z}18Z+12Z. This ideal contains 6=18⋅1+12⋅(−1)⇒18Z+12Z=6Z6 = 18 \cdot 1 + 12 \cdot (-1) \Rightarrow 18\mathbb{Z} + 12\mathbb{Z} = 6\mathbb{Z}6=18⋅1+12⋅(−1)⇒18Z+12Z=6Z
Let a,b∈Za, b \in \mathbb{Z}a,b∈Z be 2 integers. If d=gcd(a,b)⇒aZ+bZ=dZd = \gcd(a, b) \Rightarrow a\mathbb{Z} + b\mathbb{Z} = d\mathbb{Z}d=gcd(a,b)⇒aZ+bZ=dZ