Indeed, 0.999…= 1

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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: https://en.wikipedia.org/wiki/0.999…

Let’s construct Natural Numbers…

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Everybody is introduced to mathematics by “remembering” natural numbers. Today we will see, “How to construct natural numbers”.

Without much delay let’s construct “Natural Numbers”, as in “plane geometry” (you must have studied it in high school), we define “Euclid’s Axioms”, here we will define “Peano’s Axioms”, the axioms are:

• Axiom One: 1 is a natural number.
• Axiom Two: Every natural number has a successor.
• Axiom Three: 1 is not the successor of any natural number.
• Axiom Four: If the successor of $x$ equals the successor of $y$, then $x$ equals $y$.
• Axiom Five (Principle of mathematical induction): If a statement is true of 1, and if the truth of that statement for a number implies its truth for the successor of that number, then the statement is true for every natural number.

Some texts, which swap meaning of natural numbers and whole numbers, consider natural numbers to include the number zero! In that case, we will start our construction with zero instead of one (see: http://mathworld.wolfram.com/PeanosAxioms.html). This supports a very criticized philosophical idea from Indian civilization:

Everything started from nothing.

Once we are able to construct natural numbers, we can construct all kinds of number systems (with motivation of solving certain algebraic equations):

• whole numbers: add ‘0’ to the list of natural numbers (I don’t know why, but some mathematicians use whole numbers and natural numbers interchangeable)
• integers: add negative natural numbers to the list of whole numbers
• rational numbers: make fractions from integers, where denominator is not allowed to be zero.
• irrational numbers: not all numbers can be represented as fractions
• real numbers: define “Dedekind cut”, and construct real numbers out of rational numbers (lengthy task!)
• complex numbers: any polynomial equation must have a solution

There is another classification of “real numbers” , called “algebraic numbers” and “transcendental numbers“, but it is altogether a different topic of discussion which I will discuss in some other blog post.