1.10. EXISTENCE OF ROOTS 33
It is this axiom which distinguishes Calculus from Algebra. A fundamental result aboutsup and inf is the following.
Proposition 1.9.3 Let S be a nonempty set and suppose sup(S) exists. Then for everyδ > 0,
S∩ (sup(S)−δ ,sup(S)] ̸= /0.
If inf(S) exists, then for every δ > 0,
S∩ [inf(S) , inf(S)+δ ) ̸= /0.
Proof:Consider the first claim. If the indicated set equals /0, then sup(S)−δ is an upperbound for S which is smaller than sup(S) , contrary to the definition of sup(S) as the leastupper bound. In the second claim, if the indicated set equals /0, then inf(S)+δ would be alower bound which is larger than inf(S) contrary to the definition of inf(S) .
1.10 Existence of RootsWhat is 5
√7 and does it even exist? You can ask for it on your calculator and the calculator
will give you a number which multiplied by itself 5 times will yield a number which isclose to 7 but it isn’t exactly right. Why should there exist a number which works exactly?Every one you find, appears to be some sort of approximation at best. If you can’t produceone, why should you believe it is even there? Are you to accept it on faith like religion?Indeed, you must accept something without proof, but the appropriate thing to accept in thecontext of calculus is the completeness axiom of the real line on which every significanttopic in calculus depends. In calculus, roots exist because of completeness of the real lineas do integrals and all the major existence theorems in calculus, not because of algebraictechniques involving field extensions. Here is a lemma.
Lemma 1.10.1 Suppose n ∈ N and a > 0. Then if xn − a ̸= 0, there exists δ > 0 suchthat whenever y ∈ (x−δ ,x+δ ) , it follows yn −a ̸= 0 and has the same sign as xn −a.
Proof: From the binomial theorem, assuming always that |y− x|< 1,
yn −a = ((y− x)+ x)n −a =n
∑k=0
(n
k
)(y− x)n−k xk −a
=n−1
∑k=0
(n
k
)(y− x)n−k xk + xn −a = (y− x)
n−1
∑k=0
(n
k
)(y− x)n−(k+1) xk + xn −a
Now from the triangle inequality and |x− y|< 1,
(xn −a)(yn −a) = (xn −a)
((y− x)
n−1
∑k=0
(n
k
)(y− x)n−(k+1) xk +(xn −a)
)
≥=|xn−a|2
(xn −a)2 −|y− x| |xn −a|n−1
∑k=0
(n
k
)|x|k