Exams have been haunting students forever, and although you’re willing to do whatever you can to retain essential information, sometimes you end up spending weeks studying with useless revision techniques. We’re accustomed to employing our own techniques when it comes to studying such as making sticky notes, highlighting, or drawing

# Math

List of all Mathematics posts on wpgaurav

This is a continuation of the series of summer projects sponsored by department of science and technology, government of India. In this project work, I have worked to collect and expand what Ramanujan did with Nested Radicals and summarized all important facts into the one article. In the article, there

You are inside a room and there are exactly three electric bulbs outside of the room. The three bulbs have their corresponding switches (exactly three) inside the room. You can turn the switches on and off and leave them in any position. How would you identify which switch corresponds to

In an earlier post, I discussed the basic and most important aspects of Set theory, Functions and Real Number System. In the same, there was a significant discussion about the union and intersection of sets. Restating the facts again, given a collection $ \mathcal{A}$ of sets, the union of the

Just discovered Barry Martin’s Hopalong Orbits Visualizer — an excellent abstract visualization, which is rendered in 3D using Hopalong Attractor algorithm, WebGL and Mrdoob’s three.js project. Hop to the source website using your desktop browser (with WebGl and Javascript support) and enjoy the magic. PS: Hopalong Attractor Algorithm Hopalong Attractor

Sequence of real numbers A sequence of real numbers (or a real sequence) is defined as a function $ f: \mathbb{N} \to \mathbb{R}$ , where $ \mathbb{N}$ is the set of natural numbers and $ \mathbb{R}$ is the set of real numbers. Thus, $ f(n)=r_n, \ n \in \mathbb{N}, \

This mathematical fallacy is due to a simple assumption, that $ -1=\dfrac{-1}{1}=\dfrac{1}{-1}$ . Proceeding with $ \dfrac{-1}{1}=\dfrac{1}{-1}$ and taking square-roots of both sides, we get: $ \dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}$ Now, as the Euler’s constant $ i= \sqrt{-1}$ and $ \sqrt{1}=1$ , we can have $ \dfrac{i}{1}=\dfrac{1}{i} \ldots \{1 \}$ $ \Rightarrow i^2=1

These study notes on Set Theory, Functions and Real Numbers were written by Gaurav Tiwari when he was studying as a Math undergraduate in 2012-2013. The language is sought to be simple and easy to understand. Further reading material is also provided with this article. If you have any questions,