Aleena Kaiser
2022-04-25
Answered

Show that $gcd(a,b)=\left|a\right|\iff a\mid b$ ?

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gunithd5

Answered 2022-04-26
Author has **8** answers

Step 1

Assume that$a,\text{}b\in \mathbb{N}$ . I use the following definition of $D=gcd(a,b)$ :

1) D is a common divisor of a and b;

2) Every integer$d\in \mathbb{N}$ which is a common divisor of a and b divides D.

Proposition 1: If$gcd(a,b)=a$ then $a\mid b$

Proposition 2: If$a\mid b$ , then $gcd(a,b)=a$

Proof: Since$a\mid a$ and $a\mid b$ , then any integer d such that $d\mid a$ and $d\mid b$ satisfies also the condition d|a.

From$a|a,\text{}a|b$ and $d\mid a$ we conclude that $gcd(a,b)=a$

Assume that

1) D is a common divisor of a and b;

2) Every integer

Proposition 1: If

Proposition 2: If

Proof: Since

From

ophelialee4xn

Answered 2022-04-27
Author has **14** answers

Step 1

It is true that the final assertion holds if$s=0$ and $t=1$ , but it is not true that that $as+bt=az=b$ can only hold if $s=0$ and $t=1$ is false. Take, for example, $a=2$ and $b=4$ . Then $(-5)a+3\left(b\right)=b$ .

Remember that, by definition,$gcd(x,y)\mid x$ and $gcd(x,y)\mid y$ . That gives one implication.

For the converse, remember that d|x and d|y implies$d\mid gcd(x,y)$ .

It is true that the final assertion holds if

Remember that, by definition,

For the converse, remember that d|x and d|y implies

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