18 Jan 2020 - Harikrishnan N B
What is Mathematics?
Tim Poston (NIAS) summarized mathematics in two words - useful laziness. We want to be lazy and that laziness is useful. Mathematics can also be seen as a subject where Truth and Beauty meets. The beauty of mathematics is a shared experience. The useful laziness aspect can demonstrated by the following problem: As of now there are at least two people in the Lok sabha having same birthday (DD/MM). Is it a true statement or false statement?
Not only this statement is true but we can prove it without going to Lok Sabha and checking the date of birth (DD/MM) of people. In a year we have 365 days and in Lok Sabha there are 545 members so there has to be people whose birthdays coincides with other members.
This is the power of mathematics, i.e without going to Lok Sabha we can convince ourselves without any doubt that there exist atleast two people with same birthday.. That is why mathematics is seen as a useful laziness.
This is an example of {\it pigeon hole principle}. Suppose if there 365 holes and 545 pigeons, then 365 pigeons can be allocated in 365 holes and the remaining (545 - 365 = 180) pigeons can only be allocated in the already filled 365 holes. So there will exist atleast two pigeon allocated in any of the 365 holes.
Pigeonhole principle is part of a larger body of arguments called Ramsey Theory which states that- As you scale the number of mathematical objects (for eg. the number of nodes in a graph or no. of edges in a graph) then there would be certain structures that can’t be avoid. In the case of birthday problem as we increase the number of people, then we are constrained by the number of days. Hence, leading to an unavoidable structure. So, because of the shear size of the components, there is nothing like perfect randomness (since there will some structure as the number of components increases).
Another example, consider a room full of babies running around. We need to weigh the babies. Since the babies are running around, it is very difficult to weigh an individual baby. Let’s say somehow we found the average weight of babies. We also found that there is a baby who has weight very less compared to average weight. The existence of this baby proves the existence of a baby having weight very higher than the average weight.
Another example, suppose a software developer says he/she has developed a software that can compress all kinds of files losslessly. Now would you buy this software? How would you ensure whether the claim that ‘all kinds of files can be compressed losslessly’?
Let us consider a binary file of length two. According to the claim all binary files of length two can be compressed losslessly. The total number of binary files with length two are (0,0),(0,1),(1,0) and (1,1). According to the claim, there has to be one to one mapping for all the 4 possible combinations. This is because the compression is lossless. But the compression binary file with length leads to 0 and 1. The total number of binary files with length two are 4 and the total number of compressed files are 2. So there is no one to one mapping between the total number of binary files with length two and compressed file. This example shows that the claim is not True.
This argument can be extended to binary files of length \(n\). The total number of binary files with length \(n\) is \(2 ^{n}\). Now \(\forall\) \(n > 0\) and \(n \in \mathbb{Z}\), the following inequality is true:
\[\begin{equation} 2^{n} > \sum_{i=0}^{n-1}2^{i} \end{equation}\]The left hand side of the inequality is the total number of binary files of length \(n\) bit and right hand side is the total number of compressed files of size \(1, 2, \ldots, n-1\) bits. Clearly there is not a one to one mapping between the total number of binary files and total number of compressed files. Thus, the claim that all types of files can be compressed losslessly is false.
The above examples highlights the useful laziness of mathematics.
Home Work
- Write 500-1000 worded essays for each of the article (1. The Unreasonable Effectiveness of Mathematics by Richard Hamming ) and (2. You and Your Research by Richard Hamming). It can be a summary, or a response, or an inspired article based on your own experience. But, it should relate to the article in some fashion. Please write your article on the computer (MS Word format or Text format).
- In the birthday problem, if we reduce the number of people from 545 to less that 365 then what is the probability that atleast two people have the same birthday? Plot the graph of probability of collision of people having the same birthday vs. number of people.
- In the birthday problem we assumed that everyday is equally likely (uniform distribution) with respect to birthdays of people. Now instead, if we assume that the birthdays are drawn from a gaussian distribution. In this case, what is the probability that atleast two people have the same birthday? Plot the graph of probability of collision of people having the same birthday vs. number of people.
- \(\forall\) \(n, k > 0\) and \(n \in \mathbb{Z}\), is the following inequality true or false:
- \(\forall\) \(n> 0\), \(n \in \mathbb{Z}\) and \(k \in \mathbb{R}\), is the following inequality true or false: \(\begin{equation} k^{n} > \sum_{i=0}^{n-1}k^{i} \end{equation}\)
Law of Excluded Middle
In mathematics, a statement can either be true or false. It cannot be both true and false, also it cannot be neither true nor false.
There is also a set of logic which allows the possibility of both true and false with some probability. These are called paraconsistent logic.
Structure of Mathematics
- Axioms
- Rules of Inference
- Statements-Theorems, Non-Theorems and Conjectures
Proofs in Mathematics
- A direct proof is a way of showing the truth or falsehood of a given statement by a straightforward combination of established facts, usually axioms, existing lemmas and theorems, without making any further assumptions. (Wikipedia).
Statement: For any two positive integers \(a\) and \(b\) the geometric mean (GM) \(\leq\) arithmetic mean (AM). \(\begin{eqnarray} \sqrt{ab} \leq \frac{a+b}{2}\\ ab \leq \frac{(a+b)^{2}}{4}\\ 4ab \leq a^{2} +b^{2} +2ab\\ 0\leq a^{2} + b^{2} - 2ab\\ 0 \leq (a -b)^{2} \end{eqnarray}\)
\((a-b)^{2}>0\) is always true, hence for any two positive integers \(a\) and \(b\) the geometric mean (GM) \(\leq\) arithmetic mean (AM).
Statement: If \(x^{2}\) is even, then \(x\) is even, where \(x \in \mathbb{Z}\).
Lemma: If \(a\cdot b\) is even then either \(a\) or \(b\) is even.
Since \(a.b\) is even, \(a.b\) can be written in the form of \(2k\), where \(k \in \mathbb{Z}\)
Now, \(k = \frac{(a.b)}{2}\) Since \(k\) is an integer, either \(a\) or \(b\) has to be divisible by \(2\) Using the above argument, if \(x^{2}\) is even, then \(x\) has to divisible by 2, which implies \(x\) is even.
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Proof by Contradiction: Assume the contrary and stop when a contradiction is reached.
Statement: For any two positive integers \(a\) and \(b\) the geometric mean (GM) \(\leq\) arithmetic mean (AM).
Assuming the contrary : : For any two positive integers \(a\) and \(b\) the geometric mean (GM) \(>\) arithmetic mean (AM). \(\begin{eqnarray} \sqrt{ab} > \frac{a+b}{2},\\ ab > \frac{(a+b)^{2}}{4},\\ 4ab > a^{2} +b^{2} +2ab,\\ 0> a^{2} + b^{2} - 2ab\,\ 0 > (a -b)^{2}. \end{eqnarray}\)
\((a-b)^{2}<0\) is a contradiction, hence for any two positive integers \(a\) and \(b\) the geometric mean (GM) \(\leq\) arithmetic mean (AM).
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Proof by Contrapositive: If \(p => q\) then \(\sim q=> \sim p\) (\(p\) implies \(q\) then not \(q\) implies not \(p\).
For example: If there is heavy clouds then there is rain. (If there is no rain then there is no heavy clouds.)
Statement: If \(x^{2}\) is even, then \(x\) is even, where \(x \in \mathbb{Z}\).
Proving the Contrapositive statement: If \(x\) is odd then \(x^{2}\) is odd.
Any odd number $x$ can be written as \(2k + 1\), where \(k \in \mathbb{Z}\). \(\begin{eqnarray} x^{2} = (2k+1)^{2}\\ x^{2} = 4k^{2} + 4k + 1\\ x^{2} = 2k(2k + 2) + 1 \end{eqnarray}\) Let \(2k(2k + 2) = r\), where \(r \in \mathbb{Z}\) now substitute \(r\) instead of \(k(2k + 2)\). \(\begin{equation} x^{2} = 2r+ 1 \end{equation}\) From the above, \(x^{2}\) is of the form \(2r+1\), where \(r \in \mathbb{Z}\). Therefore, if \(x\) is odd, then \(x^{2}\) is also odd. Thus, we proved the contrapositive of the original statement.
- Proof by Exhaustion: is a method of mathematical proof in which the statement to be proved is split into a finite number of cases or sets of equivalent cases, and where each type of case is checked to see if the proposition in question holds. (Wikipedia).
- Computer Assisted Proof: A computer-assisted proof is a mathematical proof that has been at least partially generated by computer. (For eg. four color theorem).