Understanding another linear algebra proof

• May 11th 2013, 04:00 PM
chlovee
Understanding another linear algebra proof
Hi again!
This is a DIFFERENT question to the one I posted yesterday. This question provides less assumptions, therefore requiring different/more to prove.

Q: Assume that z $\epsilon$ F are such that a + z = a for some a $\epsilon$ F. Show that z = 0.

Solution: By assumption we have a = a + z. By one of the field axioms there exists b $\epsilon$ F such that a + b = 0, and by the commutative law 0 = b + a. Hence, using the assocative law of addition
0 = b + a = b + (a + z) = (b + a) + z = 0 + z.
By the commutative law of addition and the definition of zero: 0 = 0 + z = z + 0 = z

But why couldn't we just directly say that by the definition of 0, z + 0 = z, and by the given a + z = a, substituting a = 0 gives 0 + z = 0. Therefore combining those and using the commutative law of addition, z = z + 0 = 0 + z = 0?

​Thanks!
• May 11th 2013, 04:17 PM
Gusbob
Re: Understanding another linear algebra proof
The point is you cannot choose a. It is meant to be some element of F, not necessarily 0.