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Question 1

What is the Least Upper Bound Axiom?

Answer 1

Any non-empty set of real numbers that is bounded above has a least upper bound.

Question 2

What is Bernoulli’s Inequality?

Answer 2

(1 + x)^k ≥ 1 + kx, x > −1

Question 3

How do you prove there is a real number whose square is 2?

Answer 3

Consider the set of real numbers whose square is less than 2. Let r = Sup (S) and so either the square of r is <2, =2 or >2. Work to exclude the first and last possibility by showing that r cannot be the LUB, by taking a positive ε and finding r + ε to be in S for case 1 and r - ε to be an upper bound in case 3. This leaves only r^2 = 2.

Question 4

How do rational and irrational add and multiply.

Answer 4

Given any irrational number x and any rational number p/q, x + (p/q) and xp/q are both irrational. To see this, we argue by contradiction: if x + p/q = p′/q′ , then x = p′/q′ − p/q = (p′q − pq′)/qq′ ∈ Q, contradicting the fact that x is irrational; and similarly if x(p/q) = p′/q′ , then x = p′q/pq′ ∈ Q.

Question 5

What is the Archimedean Property?

Answer 5

For any real number r, there exists n ∈ N such that n > r.

Question 6

What is the proof of the Archimedean Property?

Answer 6

If this does not hold then there is some real number r such that n < r for all n ∈ N, i.e. the set N of all natural numbers would be bounded above. In this case, as a non-empty subset of the real numbers that is bounded above, it would have a least upper bound R. Given any n ∈ N, n + 1 is also a natural number, so for every natural number n, n + 1 ≤ R. But this implies that n ≤ R−1 for every natural number n. So R−1 is an upper bound for N, contradicting the fact that R is the least upper bound.

Question 7

How are rationals and irrationals interleaved?

Answer 7

Between any two real numbers a < b there is a rational number and an irrational number.

Question 8

Proof that rationals and irrationals are interleaved.

Answer 8

First we find a rational number between a and b.
If a < 0 < b then we can take x = 0. If the result holds for 0 ≤ a < b then it also holds for a < b ≤ 0 (find x with −b < x < −a and then take −x with a < −x < b).
So we take 0 ≤ a < b with a, b ∈ R. Choose n ∈ N such that 1/n < b − a, and let m be the smallest integer such that m > an. Then m − 1 ≤ an and an < m, which means that a < m/n ≤ a + (1/n) < a + (b − a) = b, since we choose n to satisfy 1/n < b − a. So m/n is a rational number that lies between a and b. Now we show that there is an irrational number between a and b. We use the previous result to find a rational number x such that a/√ 2 < x < b/√ 2 . We showed above that if x is rational and y is irrational then xy is irrational, so x√ 2 is irrational and a < x√ 2 < b, as required.

Question 9

What are the triangle inequalities?

Answer 9

For any two real numbers a and b,
(i) |a + b| ≤ |a| + |b| (triangle inequality);
(ii) |a − b| ≥ ||a| − |b|| (reverse triangle inequality).

Question 10

What is the definition of a convergent sequence?

Answer 10

A sequence (an) converges to a limit l ∈ R if for every ε > 0 there exists N ∈ N such that |an − l| < ε for every n ≥ N.
To show that a sequence converges, you have to show that for any ε > 0 you can find an appropriate N so that (2.1) holds.

Question 11

What is the Uniqueness of Limits Lemma?

Answer 11

A sequence can have at most one limit. This means that we can write limn→∞ an = l if an → l.

Question 12

What is the proof of the Uniqueness of Limits Lemma?

Answer 12

We suppose that an → l1 and an → l2 and show that we must have l1 = l2. Assume that l1 ≠ l2 and deduce a contradiction. Suppose that l2 > l1. Take ε = (l2 − l1)/2. Since an→l1, there exists N1 such that |an − l1| < ε for every n ≥ N1; in particular, an > l1 − (l2 − l1)/2 = (l1 + l2)/2 for all n ≥ N1. Since an → l2, there exists N2 such that |an − l2| < ε for every n ≥ N2; in particular, an < l2 + (l2 − l1)/2 = (l1 + l2)/2 for all n ≥ N2. If we take n ≥ max(N1, N2) it follows that we must have both an > (l1 + l2)/2 and an < (l1 + l2)/2, which is impossible; so we must have l1 = l2.

Question 13

What is the Shift Rule?

Answer 13

For any fixed k ∈ N, an → l as n → ∞ if and only if an+k → l as n → ∞.

Question 14

Proof of the Shift Rule?

Answer 14

Given ε > 0, choose N such that |an − l| < ε for all n ≥ N. Then, if we take n > N, we have n + k > n > N, so |an+k − l| < ε also. Conversely, given ε > 0, find M such that |am+k − l| < ε for all m ≥ M. Then if we take N = M + k we have |an − l| < ε for all n ≥ N.

Question 15

Prove that any convergent sequence is bounded?

Answer 15

Lemma 2.6. Any convergent sequence is bounded.
Proof. Take a sequence (an) such that an → a. Then, taking ε = 1 in the definition of convergence, there exists N such that |an − a| < 1 n ≥ N. So we have |an| = |(an − a) + a| ≤ 1 + |a| n ≥ N. Set A = max(|a1|, . . . , |aN−1|, |a| + 1) and then |an| ≤ A for all n ∈ N.

Question 16

Prove that if an → a then |an| → |a|.

Answer 16

Proof. Using the reverse triangle inequality ||an| − |a|| ≤ |an − a|. Given any ε > 0 choose N such that |an − a| < ε for all n ≥ N; then ||an| − |a|| < ε for all n ≥ N too.

Question 17

How does a sequence converge if it is bounded by another sequence?

Answer 17

Suppose that |an − l| ≤ |bn|, where bn → 0 as n → ∞. Then an → l.

Question 18

What are the Algebra of Limits?

Answer 18

Suppose that an → a and bn → b. Then:
(i) an + bn → a + b;
(ii) anbn → ab;
(iii) if b ≠ 0 then an/bn → a/b.

Question 19

Prove that if an ≤ bn for all n, an → a and bn → b, then a ≤ b.

Answer 19

Suppose that a > b. Choose ε = (a − b)/2; then there exists N such that for all n ≥ N, |an − a| < (a − b)/2 and |bn − b| < (a − b)/2 . So then a − an ≤ |an − a| < (a − b)/2 ⇒ an > a − (a − b)/2 = (a + b)/2 and bn − b ≤ |bn − b| < (a − b)/2 ⇒ bn < b + (a − b)/2 = (a + b)/2 . This contradicts the assumption that an ≤ bn for all n.

Question 20

What is the Sandwich Rule?

Answer 20

Suppose that an ≤ bn ≤ cn, where an → l and cn → l. Then bn → l.

Question 21

Prove the Sandwich Rule

Answer 21

Take ε > 0. Since an → l there exists N1 such that for all n ≥ N1, l − ε < an < l + ε; since cn → l there exists N2 such that for all n ≥ N2, l − ε < cn < l + ε. If we take n ≥ N := max(N1, N2) then l − ε < an ≤ bn ≤ cn < l + ε, i.e. |bn − l| < ε, which shows that bn → l.

Question 22

Prove that for any x > 0, x 1/n → 1 as n → ∞.

Answer 22

First we take x > 1 and write x^(1/n) = 1 + hn. Since x^(1/n) > 1 when x > 1, we know that hn > 0. So we have x^(1/n) = (1 + hn)^n ≥ 1 + nhn > nhn ⇒ hn < x/n . (2.3) Since we have 0 < hn < x n , it follows by the Sandwich Rule that hn → 0 as n → ∞, and so (by sum rule for limits, since x^(1/n) = 1 + hn) x^(1/n) → 1 as n → ∞. For x < 1 we can write x^(1/n) = (1/(1/x))^(1/n), and use the fact that (1/x)^(1/n) → 1 as n → ∞ and the quotient rule for limits.

Question 23

How does n^(1/n) converge?

Answer 23

n^(1/n) → 1 as n → ∞.

Question 24

What does it mean for a sequence to tend to infinity?

Answer 24

We say that an → ∞ as n → ∞ if for every R ∈ R there exists N such that an > R for every n ≥ N.

Question 25

Prove that if an → ∞ as n → ∞ then 1/an → 0 as n → ∞.

Answer 25

Given ε > 0, since an → ∞ there exists N such that an > 1/ε for all n ≥ N. Then 0 ≤ 1/an < ε for all n ≥ N, so in particular |1/an| < ε for all n ≥ N and so 1/an → 0.

Question 26

What is the Comparison Test for Limits?

Answer 26

If an ≥ bn and bn → ∞ as n → ∞, then an → ∞ as n → ∞.

Question 27

What is the Ratio Test for Limits?

Answer 27

Suppose that (an) is a sequence of positive terms, and that limn→∞ (an+1)/an = r. If r < 1 then an → 0, and if r > 1 then an → ∞. If r = 1 then we cannot deduce anything: consider the three sequences an = 1/n for each n, an = 1 for all n, and an = n for each n. In all three cases an+1/an → 1 as n → ∞. (We will see similar issues later.)

Question 28

What are the known/example limits for the Ratio Test?

Answer 28

We have the following limits as n → ∞: (i) |x^n → 0 if |x| < 1 and |x|^n → ∞ if |x| > 1; (ii) for α ∈ Q, n^α → ∞ if α > 0 and n^α → 0 if α < 0; (iii) x^n/n^k → ∞ if x > 1 (and converges to 0 if 0 ≤ x ≤ 1); (iv) x^n/n! → 0; and (v) n!/n^n → 0

Question 29

What does it mean for a sequence to be increasing, decreasing and monotonic?

Answer 29

A sequence (an) is increasing if a1 ≤ a2 ≤ a3 ≤ · · · (in general an+1 ≥ an) - the terms increase or stay the same. [If an+1 > an we say that the sequence is strictly increasing.] A sequence (an) is decreasing if a1 ≥ a2 ≥ a3 ≥ · · · (i.e. an+1 ≤ an); the terms decrease or stay the same. [If an+1 < an for all n then we say that the sequence is strictly decreasing.] A sequence is (strictly) monotonic if it is (strictly) increasing or (strictly) decreasing. [Note that a constant sequence is both ‘increasing’ and ‘decreasing’.]

Question 30

What does it mean for a sequence to be bounded?

Answer 30

A sequence (an) is bounded above if there exists M ∈ R such that an ≤ M for all n, and bounded below if there exists K ∈ R such that an ≥ K for all n. [This is the same as the set {an} being bounded above or below.]

Question 31

What is the theorem for convergence of an increasing sequence when it is/is not bounded?

Answer 31

If (an) is increasing then: (i) if it is bounded above then it converges (ii) if it is not bounded above then an → ∞.

Question 32

Prove the theorem for convergence of an increasing sequence when it is/is not bounded?

Answer 32

Consider the set A = {an : n ∈ N}. This is non-empty and bounded above (by assumption), so (by the Least Upper Bound Axiom) it has a least upper bound; we set l = sup A. We will show that an → l as n → ∞. Since l is an upper bound for all elements of A, an ≤ l for every n ∈ N. (3.1) Now take ε > 0. Since l is the least upper bound, there exists an element a of A such that a > l − ε, i.e. there exists N such that aN > l − ε. Since (an) is increasing, we know that if n ≥ N then an ≥ aN > l − ε. Combining this with (3.1) it follows that for all n ≥ N we have |an − l| < ε, which shows that an → l as n → ∞. (ii) Choose R ∈ R. Since (an) is not bounded above there exists N such that aN > R. Now since (an) is increasing, it follows that an > R for all n ≥ N; so an → ∞.

Question 33

When is a number rational?

Answer 33

A number x ∈ [0, 1] is rational if and only if its decimal expansion terminates or recurs.

Question 34

How does a subsequence of a convergent sequence converge?

Answer 34

If an → l, then any subsequence converges to l.

Question 35

Prove that any subsequence, of a convergent sequence converges, to the same limit.

Answer 35

Let (anj) be a subsequence of (an). Take ε > 0. Since an → l, there exists N such that |an −l| < ε for all n ≥ N. Now since nj → ∞, there exists J such that nj ≥ N for all j ≥ J. It follows that |anj − l| < ε for all j ≥ J, and so anj → l as j → ∞, as claimed.

Question 36

What is the Bolzano-Weierstrass Theorem?

Answer 36

Any bounded sequence of real numbers contains a convergent subsequence.

Question 37

Prove the BWT

Answer 37

If (an) is bounded then it contains a monotonic subsequence; this is bounded both above and below (since (an) is bounded) - so it is increasing and bounded above or decreasing and bounded below, so (by Theorem 3.1 or its corollary) it must converge.

Question 38

When is a sequence Cauchy?

Answer 38

A sequence (an) is Cauchy if for each ε > 0 there exists an N such that |an − am| < ε for m, n ≥ N. In fact, if a sequence is Cauchy then it must converge.

Question 39

Show that any Cauchy sequence converges?

Answer 39

Suppose that an → l. Then for any choice of ε > 0 there exists N such that |an − l| < ε/2 for all n ≥ N. Let’s take m, n ≥ N. Then |an − am| ≤ |an − l| + |l − am| < ε 2 + ε 2 = ε for all m, n ≥ N, by the triangle inequality.

Question 40

What is the General Principle of Convergence?

Answer 40

A sequence of real numbers converges if and only if it is Cauchy.

Question 41

Prove the General Principle of Convergence?

Answer 41

We have already shown that any convergent sequence is Cauchy. So suppose that (an) is a Cauchy sequence. Since (an) is bounded, it must have a convergent subsequence: anj → l as j → ∞, for some l ∈ R. We just have to show that an → l. Pick ε > 0. Since (an) is Cauchy, there exists N such that |an − am| < ε/2 for all m, n ≥ N. Since the subsequence anj → l as j → ∞, we can pick J such that nJ ≥ N and such that |anj − l| < ε/2 for all j ≥ J. If k ≥ nJ then, since nJ ≥ N, we have k, nJ ≥ N and so |ak − anJ | < ε/2; we also know that |anJ − l| < ε/2, so |ak − l| = |ak − anJ + anJ − l| ≤ |ak − anJ | + |anJ − l| < ε 2 + ε 2 = ε, which shows that ak → l as k → ∞.

Question 42

What does it mean for a series to converge?

Answer 42

If (an) is a sequence, then we say that ∑(∞/j=1) aj = s if the sequence (sn) of partial sums sn = ∑(n/j=1) aj converges to s.

Question 43

How does sum and multiplication of series work?

Answer 43

If ∑∞j=1 aj = t and ∑∞j=1 bj = s then ∑∞j=1 (aj + bj ) = t + s. If ∑∞j=1 aj = s and c ∈ R then ∑∞j=1 c*aj = cs.

Question 44

Prove that if an does not tend to 0 then ∑∞n=1 an does not converge

Answer 44

Suppose a series tends to a limit s, then by shift rule, the series up to n and n-1 both tend to s. But the value can be found by subtracting the two series, giving 0.

Question 45

How does the harmonic series converge?

Answer 45

The harmonic series is an example of a series that does not converge, despite the values in the sequences tending to 0.

Question 46

What are Vanishing Tails?

Answer 46

If ∑an converges then ∑∞j=n aj → 0 as n → ∞.

Question 47

Proof of Vanishing Tails.

Answer 47

Rearrange (4.4) as ∑∞j=kaj = ∑∞j=1 aj − ∑k-1j=1 aj , take k → ∞, using the fact that limk→∞ ∑k-1j=1 aj = ∑∞j=1aj.

Question 48

How does a series converge if its partial sums are bounded?

Answer 48

If an ≥ 0 and the partial sums ∑nj=1 aj are bounded above then ∑∞j=1 aj converges. If the partial sums are not bounded above then the series diverges.

Question 49

What is the Comparison Test for Series?

Answer 49

(Comparison Test). Suppose that 0 ≤ an ≤ bn for every n. Then * if ∑∞j=1 bj converges then ∑∞j=1 aj converges; * if ∑∞j=1 aj diverges then ∑∞j=1 bj diverges.

Question 50

e

Answer 50

We define e = ∑∞k=0 1/k! . Theorem 4.8. The number e is irrational. 4.3.2 (1 + 1/n)^n → e

Question 51

What is Absolute Convergence?

Answer 51

We say that ∑an converges absolutely if ∑|an| < ∞. There are sequences that converge that do not converge absolutely, e.g. ∑1/n does not converge whilst ∑∞k=0 (-1)k+1 1/k does converge.

Question 52

How does Absolute Convergence affect Convergence?

Answer 52

(Absolute convergence implies convergence). Suppose that ∑|an| converges. Then ∑an converges.

Question 53

Prove that Absolute Convergence affect Convergence.

Answer 53

We define two new sequences with positive terms: bn = ( an when an > 0, 0 when an ≤ 0); cn = ( 0 when an ≥ 0, −an when an < 0). Note that bn ≤ |an|, cn ≤ |an|, and an = bn − cn. Since ∑|an| converges, it follows from the Comparison Test (version 1) that both ∑bn and ∑cn converge. Since an = bn − cn, it now follows from the sum rule for series that ∑an converges.

Question 54

What is a Conditionally Convergent sequence?

Answer 54

A series that converges but does not converge absolutely (like the alternating version of the harmonic series) is called conditionally convergent.

Question 55

What is the 2nd version of the Comparison Test for series?

Answer 55

Suppose that |an| ≤ bn for every n, and ∑bn < ∞. Then ∑an converges.

Question 56

What is the Ratio Test for series?

Answer 56

(Ratio test). If an does not equal 0 and |(an+1/an)| → r as n → ∞ then: * if r < 1 the series ∑an converges absolutely; * if r > 1 the series ∑an does not converge; * if r = 1 it could be either.

Question 57

What is Cauchy's Root Test for series?

Answer 57

Suppose that |an|1/n → r as n → ∞. Then: * if r < 1 the series ∑an converges absolutely; * if r > 1 the series ∑an does not converge; * if r = 1 it could be either

Question 58

What is the Integral Test?

Answer 58

Suppose that f : [1,∞) → [0,∞) is a non-negative decreasing function. Then: * if ∫n1 f(x) dx is bounded then ∑∞n=1 f(n) converges; * if ∫n1 f(x) dx is unbounded then ∑∞n=1 f(n) diverges.

Question 59

How does a positive, decreasing sequence converge?

Answer 59

Suppose that an ≥ 0 with an+1 ≤ an and an → 0. Then ∑∞n=1 (−1)^(n+1)an converges.

Question 60

What is the definition of Continuity?

Answer 60

We say that f is continuous at c if f(x) can be made arbitrarily close to f(c) by taking x sufficiently close to c. Definition 5.1 (ε–δ definition of continuity). A function f : E → R is continuous at c ∈ E if for any ε > 0 there exists a δ > 0 such that x ∈ E and |x − c| < δ ⇒ |f(x) − f(c)| < ε.

Question 61

What is the Neighbourhood in relation to Continuity?

Answer 61

We call x ∈ E with |x−c| < δ, i.e. E ∩(c−δ, c+δ), the δ-neighbourhood of c in E.

Question 62

What is Preservation of Inequalities in relation to Continuity?

Answer 62

If f : E → R is continuous at c ∈ E and f(c) > α then there exists a δ > 0 such that f(x) > α for all x ∈ E with |x − c| < δ. Similarly, if f(c) < α then there exists a δ > 0 such that f(x) < α for all x ∈ E with |x − c| < δ.

Question 63

How does continuity hold when functions are added and multiplied?

Answer 63

If f, g : E → R are both continuous at c ∈ E then: (i) f + g is continuous at c; (ii) fg is continuous at c.

Question 64

How does continuity hold under Composition of Functions?

Answer 64

Suppose that f : E → R is continuous at c, f(E) ⊂ D, and g : D → R is continuous at f(c); then the composition of g with f, g ◦ f : E → R, defined by setting (g ◦ f)(x) = g(f(x)) x ∈ E, is continuous at c.

Question 65

How does continuity hold in relation to the modulus and maximum of a function?

Answer 65

Suppose that f, g : E → R are continuous at c ∈ E. Then: (i) the function |f| : E → R given by |f|(x) = |f(x)| is continuous at c; (ii) the function max(f, g) : E → R given by max(f, g)(x) = max(f(x), g(x)) is continuous at c.

Question 66

How do convergent sequences hold under continuity when applied to a function?

Answer 66

Suppose that f : E → R is continuous at c ∈ E, and that (xn) ∈ E is a sequence such that xn → c as n → ∞. Then f(xn) → f(c) as n → ∞.

Question 67

What is the definition of Sequential Continuity?

Answer 67

A function f : E → R is sequentially continuous at c ∈ E if whenever (xn) ∈ E and xn → c as n → ∞ then f(xn) → f(c) as n → ∞.

Question 68

What does it mean for a function to be discontinuous?

Answer 68

A function f : E → R is discontinuous at c ∈ E if there exists an ε > 0 such that for every δ > 0 there exists an x ∈ E such that |x − c| < δ but |f(x) − f(c)| ≥ ε.

Question 69

How do convergent sequences affect continuity?

Answer 69

Suppose that for any sequence (xn) ∈ E with xn → c as n → ∞ we have f(xn) → f(c) as n → ∞; then f is continuous at c.

Question 70

What are the inequalities in relation to Sin and Tan?

Answer 70

For x ∈ (0, π/2) we have: 0 < sin x < x < tan x. And for x ∈ (−π/2, π/2): |sin x| ≤ |x| ≤ |tan x|. Since |sin x| ≤ 1 the inequality |sin x| ≤ |x| is clearly valid for x ∈ (−π, π) also.

Question 71

Prove that sin is continuous?

Answer 71

Fix c ∈ R. Choose α = (x + c)/2 and β = (x − c)/2 in (8.3), which gives sin x − sin c = 2 cos((x + c)/2) sin((x − c)/2). Therefore |sin x − sin c| = 2| cos((x + c)/2)||sin((x − c)/2)| ≤ 2 × 1 × |(x − c)/2|. Continuity of f at c now follows by taking δ = ε.

Question 72

What is the Intermediate Value Theorem?

Answer 72

Suppose that f is continuous on [a, b] and that f(a) < f(b). Then for any g with f(a) < g < f(b) there exists c ∈ (a, b) such that f(c) = g. A similar statement holds if f(a) > f(b) and f(a) > g > f(b).

Question 73

When does a continuous function have a fixed point?

Answer 73

Any continuous function f : [a, b] → [a, b] has a fixed point, i.e. there exists an x∗ ∈ [a, b] such that f(x∗ ) = x∗

Question 74

Prove the fixed point from a continuous function.

Answer 74

Consider the function g(x) = f(x) − x. Then 0 ≤ g(a) and g(b) ≤ 0. If g(a) = 0 then f(a) = a, and if g(b) = 0 then f(b) = b. Otherwise g(a) > 0 and g(b) < 0, in which case by the IVT there exists a c ∈ (a, b) such that g(c) = 0, i.e. f(c) = c.

Question 75

What form can an interval be of?

Answer 75

Any interval is a set of the following form: a single point {a}, a (nontrivial) bounded interval (a, b), (a, b], [a, b), [a, b]; (9.1) a semi-infinite interval (−∞, b), (−∞, b], (a,∞), [a,∞); (9.2) or the whole line R.

Question 76

When is the range of a function an interval?

Answer 76

Let I be an interval and suppose that f : I → R is continuous. Then f(I) is an interval.

Question 77

What is the Extreme Value Theorem?

Answer 77

If f : [a, b] → R is continuous then f is bounded and attains its bounds. In other words, f is bounded above and below on [a, b], and there exist x ∗ , x∗ ∈ [a, b] such that f(x∗) = inf x∈[a,b] f(x) and f(x ∗ ) = sup x∈[a,b] f(x).

Question 78

When does an even degree polynomials have roots?

Answer 78

Let P : R → R be given by P(x) = ∑2nk akxk , a2n > 0, i.e. an even degree polynomial with positive leading coefficient. Then P has a minimum at some point x = x∗, and (i) if P(x∗) > 0 there are no real roots; (ii) if P(x∗) = 0 there is at least one real root; and (iii) if P(x∗) < 0 then there are at least two real roots.

Question 79

What is Uniform Continuity?

Answer 79

A function f : E → R is uniformly continuous on E if for any ε > 0 there exists a δ > 0 such that x ∈ E and |x − y| < δ ⇒ |f(x) − f(y)| < ε (10.2) for every x, y ∈ E. I.e. If δ only depends on ε and not on c for all c ∈ E, we say that f is uniformly continuous on E.

Question 80

How does continuity imply uniform continuity?

Answer 80

If f : [a, b] → R is continuous, then f is uniformly continuous on [a, b].

Question 81

What does it mean for a function to be increasing/decreasing?

Answer 81

A function f : E → R is said to be increasing (on E) if x ≥ y ⇒ f(x) ≥ f(y); and strictly increasing (on E) if x > y ⇒ f(x) > f(y). A function f : E → R is said to be decreasing (on E) if x ≥ y ⇒ f(x) ≤ f(y);. and strictly decreasing (on E) if x > y ⇒ f(x) < f(y).

Question 82

What is the Inverse Function Theorem for continuous functions?

Answer 82

Let I be an interval and suppose that f : I → R is continuous and strictly monotonic. Then J = f(I) is an interval and f −1 : J → I is continuous and strictly monotonic (in the same sense as f).