Is the set of even integers a field?

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A field is a set with two operations, usually called addition and multiplication, such that the set under both operations is a commutative group.

-- Commutative implies for every `a,b` in the set a+b=b+a and ab=ba.

-- In order to be a group the set with the operation must be...

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A field is a set with two operations, usually called addition and multiplication, such that the set under both operations is a commutative group.

-- Commutative implies for every `a,b` in the set a+b=b+a and ab=ba.

-- In order to be a group the set with the operation must be closed, be associative, have an identity, and have inverses. (For a field, the additive identity does not have to have a multiplicative inverse -- in other words you cannot divide by zero)

Now the even numbers are closed under addition and multiplication and are commutative and associative for both operations. There is an additive identity and each element has an additive inverse.

The problem is that there is no multiplicative identity (nor multiplicative inverses).

Therefore, the even numbers are not a field.

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A field is formed by a set of numbers which have satisfy certain conditions.

These include:

  • Closure with respect to addition to addition and multiplication: This is true for the set of even integers as for any two elements a and b, the element c = a+b and c = a*b lie in the set.
  • Associativity and commutativity for addition and multiplication holding for all elements: This is true for the set as for any three elements a,b and c we have a + (b + c) = (a + b) + c, a*(b*c) = (a*b)*c, a + b = b + a and a*b = b*a
  • Additive and multiplicative inverse lying in the set: The additive inverse of any element a or -a also lies in the set but the multiplicative inverse 1/a for every element a does not lie in the set, e.g. 1/2 does not lie in the set, though 2 does.

Therefore we cannot say that the set of even numbers is a field.

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