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198 CHAPTER 8. ALGEBRA VERSUS ANALYSIS
None of these are, as such, a function in the sense of Notation 3, though the
last is the nearest. However, trying to express x2 −1
x−1
in this notation exposes our
problems.
({( ) } )
x, x2 − 1
| x ∈ Q , Q, Q
(8.1)
x − 1
B
is illegal, because of the case x = 1. Ruling this out gives us
({( )
}
)
x, x2 − 1
| x ∈ Q \ {1} , Q \ {1}, Q . (8.2)
x − 1
B
If we regard 0 0
as ⊥, then we can have
({( ) }
)
x, x2 − 1
| x ∈ Q , Q, Q ∪ {⊥} . (8.3)
x − 1
B
Making the ⊥ explicit gives us
({( ) }
)
x, x2 − 1
| x ∈ Q \ 0 ∪ {(1, ⊥)}, Q, Q ∪ {⊥} . (8.4)
x − 1
B
If we’re going to single out a special case, we might as well write
({( ) }
)
x, x2 − 1
| x ∈ Q \ 0 ∪ {(1, 2)}, Q, Q , (8.5)
x − 1
B
dropping the ⊥, and this is equal to
({(x, x + 1) | x ∈ Q} , Q, Q) B
. (8.6)
The case of polynomials is simpler (we restrict consideration to one variable,
but this isn’t necessary). Over a ring R of characteristic zero 2 , equality of
abstract polynomials is the same as equality of the corresponding functions:
p = q in R[x] ⇔ ({(x, p(x)) | x ∈ R} , R, R) B
= ({(x, q(x)) | x ∈ R} , R, R) B
(8.7)
This is a consequence of the fact that a non-zero polynomial p − q has only a
finite number of zeros.
If we attach a meaning to elements of R(x) by analogy with (8.2), omitting
dubious points in the domain, then equality in the sense of Definition 14 is
related to the equality of Bourbakist functions in the following way
f = g in R(x) ⇔ ({(x, f(x)) | x ∈ S} , S, R) B
= ({(x, g(x)) | x ∈ S} , S, R) B
,
(8.8)
where S is the intersection of the domains of f and g.
2 See example 5 on page 20 to explain this limitation.