Proving the generator of \(\displaystyle{A}={\left\lbrace{154}{a}+{210}{b}:{a},{b}\in{\mathbb{{Z}}}\right\rbrace}\) is \(\displaystyle{\left({154},\

r1fa8dy5

r1fa8dy5

Answered question

2022-03-28

Proving the generator of
A={154a+210b:a,bZ} is (154, 210)

Answer & Explanation

cineworld93uowb

cineworld93uowb

Beginner2022-03-29Added 16 answers

Step 1
For the further proof , you just need to show that
154a+210b
just generates all the common multiples of 154 and 210 for different values of a and b where
a, bZ
and even HCF can also be written in in the form
154a+210b
for a particular choice of a and b . And HCF obviously generate all the common multiples then.

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