Monthly Archives: January 2016

Area of Rectangle


When we learn to find area enclosed by a curve, we are told to divide area in rectangular elements.


Taken from pp. 361 of Mathematics (Part – II), Class XII textbook, NCERT

But, how do we always know the value of area of rectangle? In this post, I will try to prove this well known fact in the spirit of Euclid and Cauchy.

Let’s define:

  • Boundary: A boundary is that which is an extremity of anything.
  • Figure: A figure is that which is contained by any boundary or boundaries.
  • Triangular region : A triangular region is a figure which is the union of a triangle and its interior. Also the sides of the triangle are called edges of the region and vertices of the triangle are called vertices of the region.
  • Polygonal region: A polygonal region is a plane figure which can be expressed as the
    union of a finite number of triangular regions, in such a way that if two of the
    triangular regions intersect, their intersection is an edge or a vertex of each of them.
  • Square region: It is the union of a square and its interior.

The structure in our geometry is

[\mathcal{S},\mathcal{L}, \mathcal{P}, d, m, \alpha]

where \mathcal{S} is the set of points, \mathcal{L} is the set of lines, \mathcal{P} is the set of planes, d is distance (a function satisfying first three properties of metric function), m is angular measure (a function defined for angles, with real numbers as values of the function, satisfying following postulates) and \alpha is the area function satisfying following postulates:

  1. \alpha is a function \mathcal{R} \rightarrow \mathbb{R}, where \mathcal{R} is the set of all polygonal regions and \mathbb{R}  is the set of all real numbers.
  2. For every polygonal region R, \alpha(R) > 0.
  3. If two triangular regions are congruent, then they have the same area.
  4. If two polygonal regions intersect only in edges and vertices (or do not intersect at all), then the area of their union is the sum of their areas.
  5. If a square region has edges of length 1, then its area is 1.

Proposition 1. If a square has edges of length 1/q (q a positive integer), then its area is 1/q^2.

Proof. A unit square region can be decomposed into q^2 square regions, all with the same edge 1/q as


Then all smaller square have the same area A (divide each square into triangles using diagonals and then use Postulate 3 to prove that all of them have same area). Therefore 1 = q^2A (from Postulate 4) and A = 1/q^2.

Proposition 2. If a square has edges of rational length p/q, then its area is p^2/q^2.

Proof. Such a square can be decomposed into p^2 squares, each of edge 1/q as:


If A is its area, then

A = p^2 \times \frac{1}{q^2} = \frac{p^2}{q^2}

Proposition 3.[Peter Lawes] If a square has edges of length a, then its area is a^2.

Proof. Given a square S_a with edges of length a. Given any rational number p/q,  let S_{p/q} be a square of edge p/q, with an angle in common with S_a, as:


Then, \frac{p}{q} < a hence S_{p/q} lies inside S_a. For some real number s (by using Postulate 4) we get:

\alpha(S_{p/q}) + s = \alpha(S_{a})

\Rightarrow \alpha(S_{p/q}) < \alpha(S_{a})

\Rightarrow \frac{p^2}{q^2}< \alpha(S_{a})

\Rightarrow\frac{p}{q} < \sqrt{\alpha(S_{a})}

But, selection of \frac{p}{q} being arbitrary, the upper-bound should be unique. Since there exists a unique supremum of the set consisting all possible side lengths of smaller square in \mathbb{R}, we can claim:

a = \sqrt{\alpha(S_{a})}

We can prove this claim by following the proof of statement: \sup\{x \in \mathbb{R} : 0 \leq x, x^2 < 2\} = \sqrt{2}. Hence:


Theorem: Area of rectangle is equal to the product of length of any two adjacent sides.

Proof. Given a rectangle of base b and altitude h, we construct a square of edge b + h, and decompose it into squares and rectangles as:

Then from Postulate 4, we get:

(b+h)^2 = 2A + A_1 +A_2

b^2 + 2bh + h^2=2A +h^2+b^2

2bh = 2A

bh = A



Moise, Edwin (1990). Elementary Geometry from an Advanced Standpoint. Addison-Wesley Pub. Co.

Thanks to Dr. Shailesh Shirali for pointing out this beautiful book.


The World of Exponents


We all use calculators, spreadsheets, etc. for doing various kinds of calculations like addition, multiplication, division and subtraction. Now let us consider the operation of multiplication. When we multiply same number we are able to represent it in terms of exponents. For example:

2 \times 2 = 2^2
Thus exponents naturally creep into our mathematical notations, because of their short-hand notation and convenience. Now let us consider exponents as an independent operation (like we created multiplication out of repeated addition). Thus:
a^n = \underbrace{a \times a \times \ldots \times a}_\text{n times} \qquad \text{where,} \quad n \quad \text{is a natural number \&} \quad a \quad \text{is an integer.}

Further we say that:

a^{-n} = \underbrace{\frac{1}{a} \times \frac{1}{a} \times \ldots \times \frac{1}{a}}_\text{n times} \qquad \text{where,}\quad n \quad \text{is a natural number \&} \quad a \quad \text{is a non-zero integer.}

Now after handling negative and positive integers we can define for zero:

a^0 = 1 \qquad \text{where,}\quad a \quad \text{is a non-zero integer.}
0^n = 0 \qquad \text{where,}\quad n \quad \text{is a positive integer.}
\text{in fact} \quad 0^n = \text{Not Defined} = \infty \quad \text{if n is a negative integer}

Here I am leaving one case, namely 0^0 which has a very long and controversial history. We will come back to it after some time.

Now since we know other Number Systems let us try to generalize this exponent notation into what we call exponential function as:

f(x) = a^x \qquad \text{where, a is a fixed non-negative real number and x is any real number}

Now let us consider few examples to understand this function; fix a=2, then:

f(2) = 2^2 = 4
f(3) = 2^3 = 8
f(1.5) = 2^{1.5} = 2^{\frac{3}{2}} = 2\sqrt{2}
f(1.765) = 2^{1.765} = ?
f(\sqrt{2}) = 2^{\sqrt{2}} = ??

But if we play with base i.e. a, it isn’t that much interesting for me, for example, set a=\sqrt{2} this is equivalent to a=2^{\frac{1}{2}}

But calculators can calculate all these values easily. HOW? They have inbuilt logarithmic tables!!

Now let us return to our most weird case of generalization:

What is the value of 0^0?

This is a very old question and many big-shots have tried to answer this question, and here I try to provide a big-picture of their attempts:

  •  Euler argues for 0^0 = 1 since a^0 = 1 for a\neq 0. The controversy raged throughout the nineteenth century.
  • Cauchy had listed 0^0 together with other expressions like \frac{0}{0} and \infty - \infty in a table of undefined forms.

But in book Concrete Mathematics [p.162] authors remarked:

We must define x^0 = 1 for all x, if the binomial theorem is to be valid when x = 0, y = 0 and x = -y. The theorem is too important to be arbitrarily restricted. By contrast, the function 0^x is quite unimportant.

But Alex Lopez-Ortiz firmly says:

But no, no, ten thousand times no! Anybody who wants the binomial theorem (x + y)^n = \sum_{k = 0}^n \binom{n}{k} x^k y^{n-k} to hold for at least one non-negative integer n must believe that 0^0 = 1, for we can plug in x = 0 and y = 1 to get 1 on the left and 0^0 on the right.


Indeed, 0.999…= 1


There is a video by Vi Hart titled “Why Every Proof that .999… = 1 is Wrong”

It is true that few real numbers (with cardinality same as that of rational numbers) can’t have unique decimal representation, like, 0.493499999… = 0.4935000…. So, the point made in this video is that you can’t prove two different representations of a number to be equal.

But why so? As pointed out in my earlier post, this has something to do with the way we construct numbers. Such ambiguity in representation holds irrespective of representation system (binary or decimal)

In decimal representation of real numbers  we subdivide intervals into ten equal subintervals. Thus, given x\in [0,1], if we subdivide [0,1] into ten equal subintervals, then x belongs to a subinterval \left[\frac{b_1}{10}, \frac{b_1 + 1}{10}\right] for some integer b_1 in \{0,1,\ldots ,9\}. We obtain a sequence \{b_n\}_n of integers with 0 \leq b_n \leq 9 for all n \in \mathbb{N} such that x satisfies

\frac{b_1}{10}+ \frac{b_2}{10^2}+ \ldots + \frac{b_n}{10^n}\leq x \leq \frac{b_1}{10} + \frac{b_2}{10^2} + \ldots + \frac{b_n+1}{10^n}

In this case we say that x has a decimal representation given by

x = 0.b_1 b_2\ldots b_n \ldots

The decimal representation of x\in [0,1] is unique except when x  is a subdivision point at some stage, which is when x=\frac{m}{10^n} for some m,n \in \mathbb{N}; 1 \leq m \leq 10^n. We may also assume that m is not divisible by 10.

When x is a subdivision point at the n^{th} stage, one choice for b_n corresponds to selecting the left subinterval, which causes all subsequent digits to be 9, and the other choice corresponds to selecting the right subinterval, which causes all subsequent digits to be 0.

For example, x = \frac{1}{2} = \frac{5}{10} = 0.4999\ldots = 0.5000\ldots unlike x = \frac{1}{3} = 0.333...

A nice exposition is available on Wikipedia:…

Little Things Matter


This is a very famous puzzle.  Here is motivation:

Infinite Chocolate (gif from

Now consider following variation.


Where does the hole in second triangle come from? (image credit: Rookie1Ja ,

The 64 = 65 paradox arises from the fact that the edges of the four pieces, which lie along the diagonal of the formed rectangle, do not coincide exactly in direction. This diagonal is not a straight segment line but a small lozenge (diamond-shaped figure), whose acute angle is

\arctan(\frac{2}{3}) - \arctan( \frac{3}{8}) = \arctan (\frac{1}{46})

which is less than 1 degree 15′ . Only a very precise drawing can enable us to distinguish such a small angle. Using analytic geometry or trigonometry, we can easily prove that the area of the “hidden” lozenge is equal to that of a small square of the chessboard.


It looks like a triangle, because a thick line was used. Hypotenuse of the composite triangle is actually not a straight line – it is made of two lines. Forth cusps are where the arrows point (c9, l6).

Also there is an interesting video illustrating this in real life: