Smart Fallacies: i=1, 1= 2 and 1= 3

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 \ldots \{2 \}$ . This

Set Theory, Functions and Real Numbers

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, feel free to send a

Free Online Calculus Text Books

In this list I have collected all useful and important free online calculus textbooks mostly in downloadable pdf format. Feel free to download and use these. Elementary Calculus : An approach using infinitesimals by H. J. Keisler Multivariable Calculus by Jim Herod and George Cain Calculus by Gilbert Strang Calculus Bible by

Solving Ramanujan’s Puzzling Problem

Consider a sequence of functions as follows:- $ f_1 (x) = \sqrt {1+\sqrt {x} } $ $ f_2 (x) = \sqrt{1+ \sqrt {1+2 \sqrt {x} } } $ $ f_3 (x) = \sqrt {1+ \sqrt {1+2 \sqrt {1+3 \sqrt {x} } } } $ ……and so on to $ f_n (x) = \sqrt {1+\sqrt{1+2 \sqrt