MHB What is the value of gcd(0,0) in mathematics?

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The value of gcd(0,0) is debated, with some arguing it should be defined as 0 by convention, while others suggest it could be 1 since every integer divides 0. However, defining gcd(0,0) as 1 would violate mathematical properties, such as the relationship between gcd and multiplication. Many mathematical texts assume at least one of the numbers is nonzero in their definitions, leading to the conclusion that gcd(0,0) is undefined or infinite. The discussion highlights the importance of conventions in mathematics to avoid special cases in theorems. Ultimately, the consensus leans towards gcd(0,0) being conventionally considered as 0.
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I have read that by convention gcd(0,0)=0. But surely 1 fits the bill. Everything divides 0, including 1. But since 1 divides everything we must have gcd(0,0)=1
 
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Poirot said:
I have read that by convention gcd(0,0)=0. But surely 1 fits the bill. Everything divides 0, including 1. But since 1 divides everything we must have gcd(0,0)=1

What's your source for the convention that gcd(0,0) = 0? I looked at several books on number theory and they all include in the definition of gcd(a,b) the assumption that at least one of a and b is nonzero. Besides, since every positive integer divides 0, there is no greatest common divisor of 0 and 0.
 
What's your source for the convention that gcd(0,0) = 0? I looked at several books on number theory and they all include in the definition of gcd(a,b) the assumption that at least one of a and b is nonzero. Besides, since every positive integer divides 0, there is no greatest common divisor of 0 and 0.
No, it is a convention: see here. Having $\gcd{(0, 0)} = 1$ would break the following property:

$$m \cdot \gcd{(a, b)} = \gcd{(m \cdot a, m \cdot b)} ~ ~ ~ \text{for} ~ m \geq 0$$

Which would yield $1 = \text{anything}$.

Generally, conventions like these are chosen to minimize the amount of special cases theorems have to deal with. In and of themselves, they are rather trivial - in practice, it's not very useful to know whether $\gcd{(0, 0)} = 0 ~ \text{or} ~ 1$.
 
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Bacterius said:
No, it is a convention: see here. Having $\gcd{(0, 0)} = 1$ would break the following property:

$$m \cdot \gcd{(a, b)} = \gcd{(m \cdot a, m \cdot b)} ~ ~ ~ \text{for} ~ m \geq 0$$

Which would yield $1 = \text{anything}$.

Generally, conventions like these are chosen to minimize the amount of special cases theorems have to deal with. In and of themselves, they are rather trivial - in practice, it's not very useful to know whether $\gcd{(0, 0)} = 0 ~ \text{or} ~ 1$.

I do not consider Wolfram|Alpha a good reference.
There are many cases where it does not give the proper mathematical answer.
I don't blame it - it's a calculator and not a math reference.

As for gcd(0,0), I don't see specific references to it on wiki or on wolfram|mathworld.
However, both give the definition that it "is the largest positive integer that divides the numbers without a remainder."
Since any large positive integer divides 0, it would follow that gcd(0,0) is undefined (or infinity).
 
I like Serena said:
I do not consider Wolfram|Alpha a good reference.
There are many cases where it does not give the proper mathematical answer.
I don't blame it - it's a calculator and not a math reference.

As for gcd(0,0), I don't see specific references to it on wiki or on wolfram|mathworld.
However, both give the definition that it "is the largest positive integer that divides the numbers without a remainder."
Since any large positive integer divides 0, it would follow that gcd(0,0) is undefined (or infinity).


The assumption was that the people who wrote Wolfram|Alpha probably know their stuff, and would correctly handle the special cases. Of course it's not an official reference, but it's handy to quickly check things :)
 
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