Back in my school days, my mathematics professor always used to praise maths as Natural Science. It is indeed. However during school / college days, I used to wonder why the hell I’m learning the differentiation / integration / ordinary differential equations / Fourier transforms bla bla bla. However, after spending 7 odd years in IT career and filling by mind with arrays, structures, sockets, RPCs, observer design patters, memory leaks, core dumps, GDB, conference calls, proposals, delivery Huhhhhh 😦 I wanted to go back to my 10 std and immerse myself in trigonometry & algebra.

My friends used to mock me being a nerd as I always use to tell speed of the light is not \(3 \times 10^8 m/sec\) , it is \(2.99792458 \times 10^6 m/sec\) . :) [ Now I know, why those people called me a nerd :) ]

Mathematics & Physics were my most sought-after subject. It still intrigues me when I see movies like A beautiful mind Apollo 13 Inter-Stellar Martian etc. Not just for sci-fi scenes ( could anyone forget Back to the future series wow… ), but for the subtle physical & mathematical facts that is buried deep inside story and dialogues.

Of all the mathematical figures, \(\sqrt{2}\) is my favorite number. It is because, it a number, however it is categorized as irrational.

Irrational means, not being rational. i.e going bonkers

Well let me explain how.

Any rational number can be explained in fractional form. So let us say \(\sqrt{2}\) can be expressed as by 2 numbers \(p\) & \(q\), such that, \(\frac {p}{q} = \sqrt{2}\)

But we should also remember that, P or Q should be ODD number. Because, if both P & Q are even number, then should get cancelled until, they become odd. Just think about it, \(\frac {128}{48}\) will eventually become \(\frac {8}{3}\) . So our first premise is this.

\(\sqrt{2}\) is a rational number expressed by \(\frac {p}{q} = \sqrt{2}\), such that P or Q are odd numbers.

Now let us try to prove our premise and try to add support for our argument.

\[\begin{align} \sqrt{2} & = \frac{p}{q} \\ & = \frac{2n}{q} \text{p = 2n} \\ & = 2 = \frac{4n^2}{q^2} \text{squaring on both sides} \\ & = 2q^2 = 4n^2 \text{re-arranging} \\ & = q^2 = 2n^2 \text{Taking out the common factor} & = \approx q^2 \approx \text{q is even} \end{align}\]

Eq-1

$$ \frac {p}{q} = \sqrt{2} $$ [ premise ]

Eq-2

$$ \frac {p^2}{q^2} = 2 $$ [ squaring on both sides ]

Eq-3

$$ p^2 = 2q^2 $$ [ re-arranging ]

Since anything multiplied by \(2\) is even number, we can term that, \(p^2\) is even . If \(p^2\) is even, then even \(p\) is even, since square of an odd number is odd and square of even number is even.

Okay, so now we had deduced \(p\) is even, there can be only one thing possible to prove the above premise. i.e \(q\) is Odd. Let us try to find out, whether \(q\) is odd or not.

Eq-4 \(\frac {2n}{q} = \sqrt{2}\) [ Given p is even, any even number p can be represented as \(p = 2n\), where n being odd or even ]

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