Proof that gcd(a, 0) = |a|, gcd(a, a) = |a|, and gcd(a, 1) = 1

[Math100]

Homework Statement

For a nonzero integer a, show that gcd(a, 0)=abs(a), gcd(a, a)=abs(a), and gcd(a, 1)=1.

Relevant Equations

None.

Proof: First, we will show that gcd(a, 0)=abs(a).
Suppose a is a nonzero integer such that a$\neq$0.
Note that gcd(a, 0)$\le$abs(a) by definition of the greatest common divisor.
Since abs(a) divides both a and 0,
we have that gcd(a, 0)=abs(a).
Therefore, for a nonzero integer a,
we have shown that gcd(a, 0)=abs(a).
Next, we will show that gcd(a, a)=abs(a).
Suppose a is a nonzero integer such that a$\neq$0.
Note that gcd(a, a)$\le$abs(a) by definition of the greatest common divisor.
Since abs(a) divides a, we have that gcd(a, a)=abs(a).
Therefore, for a nonzero integer a,
we have shown that gcd(a, a)=abs(a).
Finally, we will show that gcd(a, 1)=1.
Suppose a is a nonzero integer such that a$\neq$0.
Note that gcd(a, 1)$\le$1 by definition of the greatest common divisor.
Since 1 divides both a and 1,
we have that gcd(a, 1)=1.
Therefore, for a nonzero integer a,
we have shown that gcd(a, 1)=1.

 

[PeroK]

Next, we will show that gcd(a, a)=abs(a).
Suppose a is a nonzero integer such that a$\neq$0.
Note that gcd(a, a)$\le$abs(a) by definition of the greatest common divisor.

Are you sure you are using the definition of GCD there and not some basic property that can be proved from the definition?

 

[Math100]

Are you sure you are using the definition of GCD there and not some basic property that can be proved from the definition?

No, I am not sure. What's the definition of GCD then?

 

[PeroK]

No, I am not sure. What's the definition of GCD then?

https://en.wikipedia.org/wiki/Greatest_common_divisor#Definition

 

[Math100]

First, we will show that gcd(a, 0)=abs(a).
Suppose gcd(a, 0)=d, where d is an integer and a is a nonzero integer.
Note that d divides both a and 0.
Since each nonzero integer divides 0,
it follows that d is the largest divisor of a.
Thus, we have that gcd(a, 0)=abs(a).

Is this right for the first subproof of this problem?

 

[PeroK]

First, we will show that gcd(a, 0)=abs(a).
Suppose gcd(a, 0)=d, where d is an integer and a is a nonzero integer.
Note that d divides both a and 0.
Since each nonzero integer divides 0,
it follows that d is the largest divisor of a.
Thus, we have that gcd(a, 0)=abs(a).

Is this right for the first subproof of this problem?

I don't see that you tackle the only real issue here which is why we can't have $gcd(a, 0) > |a|$.