Calculus

Prameya

City of Lights

May 26, 2025

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“Indeed there is no royal road through mathematics, but we do not need to break up the asphalt and
destroy the signage to make travelling what roads there are a trial of one’s skill.” — Jordan Bell.

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 CONTENTS

1 The Least Upper Bound Property 1

1.1 First Problem with Rationals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Defect in Rationals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.3 Least Upper Bound Property of Reals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.4 Archimedean Property of Reals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.5 Dense Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.6 Surds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.7 Dirichlet’s Approximation Theorem (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Sequences 7

3 Solutions to Some Exercises 9

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iv CONTENTS

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 CHAPTER 1
THE LEAST UPPER BOUND PROPERTY

1.1 FIRST PROBLEM WITH RATIONALS

Exercise 1.1. Show that there is no rational number x such that x2 = 2.a
a This
√ √
is usually stated by saying that 2 is not rational. But, as of now, we do not know that 2 exists.

1.2 DEFECT IN RATIONALS

§ Definition 1.1 Let S be non-empty subset of Q. We say that a ∈ Q is a lower bound of S if

a ≤ s for all s ∈ S (1.1)

Similarly, we say b ∈ Q is an upper bound of S if

s ≤ b for all s ∈ S (1.2)

We say S has a lower bound in Q if there is a ∈ Q such that a is a lower bound of S, and, similarly, we
say S has an upper bound in Q if there is b ∈ Q such that b is an upper bound of S. When S has a lower
bound in Q then we also say that S is bounded below in Q. Similarly, when S has an upper bound in Q
then we also say that S is bounded above in Q.

Example 1.1. Following are some simple examples.

a) The set of all the natural numbers does not have an upper bound in Q. However, it does have a lower bound.
One such lower bound is 1.
b) The set {1/n : n ∈ N} is both bounded above and below in Q.

✍ 1.2 Defect in Rationals. Let

S = {x ∈ Q : x > 0, x2 ≥ 2} (1.3)
Note that S has a lower bound in Q. Let L denote the set of all rational numbers which are lower bounds to S. We
show that L does not have a maximum. More precisely, we show that there is no element ℓ in L such that a ≤ ℓ
for all a ∈ L. 1 Let a ∈ L be arbitrary. There are two cases.
Case 1. a ∈ S.
1 We
√
cannot use the symbol ‘ 2’ anywhere in the proofs. Our proof should stay in the rationals. In other words, we have to pretend
for the moment that we only know about the rationals and not about the reals.

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2 CHAPTER 1. THE LEAST UPPER BOUND PROPERTY

In this case a2 > 2 and let δ = a2 − 2. Define an = a − 1/n for each natural number n. Now

2a 1 2a 2a
a2n = a2 − + 2 > a2 − =2+δ− (1.4)
n n n n

Choose a natural number N such that N δ > 2a.2 By the above calculation, we see that a2N > 2, and hence aN ∈ S.
But a is, in particular, a lower bound of S, and hence a < aN , but this is contradiction.
Case 2. a ∈ / S.
In this case a2 < 2 and let 2 − a2 = δ. Define an = a + 1/n for all n ∈ N. An argument similar to the one given
in the previous case shows that there is N such that aN is also a lower bound of S. Since aN > a, we have shown
that a is not the greatest lower bound of S, and we are done. ♢

✍ 1.3 Takeaway. By Point 1.2 we see that there exist subsets of Q which have a lower bound in Q but do not
have a greatest lower bound in Q. Similarly, we can construct subsets of Q which have an upper bound in Q but
not having a least upper bound. This should invoke the feeling that there are ‘holes’ in the rationals. ♢

Exercise 1.2. Let S be a subset of Q. If b ∈ Q is an upper bound of S, show that each rational that exceeds b is also
an upper bound of S. Prove an analogous results about lower bounds.

1.3 LEAST UPPER BOUND PROPERTY OF REALS

§ Definition 1.2 Let S be non-empty subset of R. We say that a ∈ R is a lower bound of S if

a ≤ s for all s ∈ S (1.5)

Similarly, we say b ∈ R is an upper bound of S if

s ≤ b for all s ∈ S (1.6)

We say S has a lower bound in R if there is a ∈ R such that a is a lower bound of S, and, similarly, we say
S has an upper bound in R if there is b ∈ R such that b is an upper bound of S.

From now on ‘(greatest) lower bound’ would mean ‘(greatest) lower bound in R’ and similarly for ‘(least) upper
bound.’

✍ 1.4 Least upper bound property of the Reals (Axiom.) The real numbers have the least upper bound
property. More precisely, if a subset S of R has an upper bound, then the set of all the upper bounds of S has a
smallest element. This element is called the least upper bound of S. ♢

✍ 1.5 Notation. Let S be a subset of R. If S is bounded above then the least upper bound of S is denoted by
sup(S). In symbols,
sup(S) = min{b ∈ R : s ≤ b for all b ∈ S} (1.7)
If S does not have an upper bound then we denote this by writing

sup(S) = ∞ (1.8)

We refer to sup(S) as the supremum of S. Similarly, is S has a lower bound then the greatest lower bound of S
is denotes by inf(S). In symbols,

inf(S) = max{a ∈ R : a ≤ s for all s ∈ R} (1.9)

2 Such an N exists for the following reason. Let δ/a = p/q, where p and q positive integers. Then N = 3q does the job.

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1.3. LEAST UPPER BOUND PROPERTY OF REALS 3

If S does not have a lower bound that we denote this by writing

inf(S) = −∞ (1.10)

We refer to inf(S) as the infimum of S. ♢

Theorem 1.6. We have  

1
inf : n∈N =0 (1.11)
n

Proof . Let S = {1/n : n ∈ N}. Clearly, 0 is a lower bound of S and hence inf(S) ≥ 0. Let
 
1
T = : n∈N (1.12)
2n
Write s to denote inf(S) and t to denote inf(T ). We want to show that s = 0. Since T ⊆ S, we have t ≥ s. Now
1 1
t≤ for all n ∈ N ⇒ 2t ≤ for all n ∈ N (1.13)
2n n
Thus 2t is a lower bound for S. But s is the greatest lower bound of S, and hence 2t ≤ s. This along with the
previous deduction that t ≥ s shows that s cannot be positive. So the only possibility for s is that s = 0. ■

Exercise 1.3. Let S be a subset of R. Assume that S has a lower bound. Show that S also has a greatest lower
bound.

Exercise 1.4. Let S and T be subsets of R. Assume T ⊆ S. Show that

inf(T ) ≥ inf(S) and sup(S) ≥ sup(T )

Here we follow the convention that a ≤ ∞ for all a ∈ R as well as a = ∞ and similarly for the analogous thing for
‘−∞.’

Exercise 1.5. Let A ⊆ R be non-empty. Suppose x ∈ R is such that a ≤ x for all a ∈ A. Show that sup(A) ⩽ x.

Exercise 1.6. Let A and B be non-empty subsets of R such that a ⩽ b for all a ∈ A and all b ∈ B. Show that
sup A ≤ inf B.

Exercise 1.7. Show that for every a > 0 we have

∞ 
\ ai
0, =∅ (1.14)
n=1
n

Exercise 1.8. Let A and B be non-empty subsets of R with both A and B bounded above. Define

A + B = {a + b : a ∈ A, b ∈ B}

Show that sup(A + B) = sup A + sup B.

Exercise 1.9. Let A, B ⊆ [0, ∞) be non-empty. Assume that both A and B are bounded above. Define

AB = {ab : a ∈ A, b ∈ B}

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4 CHAPTER 1. THE LEAST UPPER BOUND PROPERTY

Show that sup(AB) = (sup A) · (sup B).

Exercise 1.10. Let S be a non-empty set consisting of positive reals. If inf(S) = 0, the show that for each n ≥ 1
there is s ∈ S such that s < 1/n.

1.4 ARCHIMEDEAN PROPERTY OF REALS

Theorem 1.7. Archimedean Property of Reals. Let a be a positive real and b be any real number.
Then there is a natural number n such that na > b.
Proof . If b ≤ 0 then there is nothing to do, so we may assume that b is positive. Let
S = {1/n : n ∈ N}
We know that inf(S) = 0, and hence a is not a lower bound of S. Consequently, there is a natural number n such
that 1/n < a/b and we are done. ■

1.5 DENSE SETS

§ Definition 1.3 A subset A of R is said to be dense in R if A intersects each open interval. More precisely,
we say that A is dense in R if whenever x and y are real numbers such that x < y, there is a point a ∈ A such
that x < a < y.

Example 1.8. Note the following.

a) The integers are not dense in R since there is no integer in the open interval (1/2, 3/4).
b) If A is dense in R then every superset of A is also dense in R.
c) R is dense in R.

Theorem 1.9. The set of all the rational numbers is dense in R.

Proof . Let x and y be two real numbers such that x < y. We want to show that there is a rational number
between x and y. If x and y have opposite sign then 0 would be one such rational number. So we may assume that
x and y has the same sign. For concreteness, assume that x and y are both positive.
The idea is very simple. We choose a large n and start at the origin, taking a step size of 1/n each time.
Eventually we would fall in the pit that is the open interval (x, y). We now execute this intuition. By the
Archimedean property of reals (or by Theorem 1.6) we know that there is a natural number N such that 1/N < y−x.
If 1/N is already in (x, y) then we are done. So we may assume that 1/N is not in (x, y). It follows that 1/N ≤ x.
Define  
k
S= k∈N: ≤x (1.15)
N
Note that by the Archimedean property of reals the set S is finite. Also, since 1 is in S, the set S is a non-empty.
So S is a finite subset of N, and hence has a maximum element, which we will denote by M . So M is in S and
M + 1 is not in S, and hence
M M +1
≤ x, >x (1.16)
N N
But
M M 1 1 M +1 M +1
≤x ⇒ + ≤x+ ⇒ < x + (y − x) ⇒ x< <y (1.17)
N N N N N N
and we are done. ■

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1.6. SURDS 5

Exercise 1.11. Let A be dense subset of R. Show that for any x ∈ R the set {x + a : a ∈ A} is dense in R. Also
show that for any nonzero x ∈ R the set {xa : a ∈ A} is dense in R.

Exercise 1.12. Is it necessary that the intersection of two dense subsets of R is also dense in R.

Exercise 1.13. Let A be a dense subset of R. Show that A intersects every open interval in infinitely many points.

1.6 SURDS
✍ 1.10 Square roots of positive reals. We show that there is a positive real number x0 such that x20 = 2. The
method of proof can be used to show the existence of square roots of all positive reals. Define the set

S = {x ∈ R : x2 < 2} (1.18)

Then S has an upper bound. Let x0 = sup(S) and note that x0 > 0. We claim that x20 = 2. Suppose not. Then
there are two cases; either x20 < 2 or x20 > 2. Suppose first that x20 < 2. Say δ = 2 − x20 and note that δ > 0. Then
define, for each n ∈ N, xn = x0 + 1/n. Then
   
1 2x0 1 2x0 1 2x0 2x0 + 1
x2n = x20 + 2 + ⇒ 2 − x2n = (2 − x20 ) − 2
+ = δ − 2
+ >δ− (1.19)
n n n n n n n

By the Archimedean property of reals we can find a natural number N such that δ > (2x0 + 1)/N . Thus, by the
above calculation, we see that 2 − x2N > 0, and hence xN is in S. But xN > x0 , contradicting the fact that x0 is an
upper bound of S.
Now suppose x20 > 2. By a reasoning similar to the above, we can show that there is N such that x0 − 1/N also
satisfies x2N > 2, and also that xN > 0. It then follows that xN is also an upper bound of S. But now x0 > xN
contradicts the fact that x0 is the least upper bound of S. This finishes the proof. This also shows that the set of
all the reals is strictly bigger than the set of rationals. ♢

§ Definition 1.4 A real number is said to be irrational if it is not rational.

Theorem 1.11. Let n be a positive integer. Then for each non-negative

√ real number x, there is a unique
non-negative real number y such that y n = x. We denote this y as n x or at x1/n .

Proof . Exercise. ■

Exercise 1.14. Show that the set of all the irrational real numbers is dense in R.

√ √
Exercise 1.15. Show that 2 + 3 is irrational.

√ √ √
Exercise 1.16. Show that 2 + 3 + 5 is irrational.

Exercise 1.17. Let α, β and γ be rational numbers such that

√ √ √
α 2+β 3+γ 5=0 (1.20)

Show that α = β = γ = 0.

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6 CHAPTER 1. THE LEAST UPPER BOUND PROPERTY

Exercise 1.18. Is it true that the sum of two irrational numbers is irrational?

1.7 DIRICHLET’S APPROXIMATION THEOREM (OPTIONAL)

Theorem 1.12. Dirichlet’s Approximation Theorem. Let x be an irrational number. Then for each
natural number N there is a positive integer q such that

p 1
x− < (1.21)
q qN

for some integer p.

Proof . Let {y} denote the fractional part of any real number y. Let N be a natural number. Consider the N
numbers
{x}, {2x}, {3x}, . . . , {N x} (1.22)
and let  
k k+1
Sk = ,
n n
for k = 0, 1, . . . , N − 1. If there is q ∈ {1, . . . , N } such that {qx} ∈ S0 , then we would have

⌊qx⌋ 1
0 < {qx} < 1/N ⇒ 0 < qx − ⌊qx⌋ < 1/N ⇒ x− < (1.23)
q qN

and we are done since q ≤ N . So we may assume that none none of the {qx} lie in S0 . But then, by the pigeonhole
principle, there must exists a, b ∈ {1, . . . , N } such that a < b and {ax}, {bx} ∈ Sk for some k ≥ 1. Thus

1 1 ⌊ax⌋ − ⌊bx⌋) 1
|{bx} − {ax}| < ⇒ |(b − a)x + (⌊ax⌋ − ⌊bx⌋)| < ⇒ x− < (1.24)
N N b−a (b − a)N

and we are done since b − a < N . ■

✍ 1.13 Takeaway. The above theorem is a quantitative version of the fact that the rationals are dense in R. It
tells us that in order to approximate a given irrational number by a rational number up to a tolerance of 1/N , one
does not need to look for rationals with very big denominators. We are guaranteed to find a rational which, when
expressed in lowest terms, has a denominator at most N . ♢

Exercise 1.19. Let √

S = {m + n 2 : m, n ∈ Z} (1.25)
Show that S is dense in R.

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 CHAPTER 2
SEQUENCES

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8 CHAPTER 2. SEQUENCES

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 CHAPTER 3
SOLUTIONS TO SOME EXERCISES

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10 CHAPTER 3. SOLUTIONS TO SOME EXERCISES

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 BIBLIOGRAPHY

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