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

# Functions

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,

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 https://www.math.wisc.edu/~keisler/keislercalc-12-23-18.pdf Multivariable Calculus by Jim Herod and George Cain http://people.math.gatech.edu/~cain/notes/calculus.html Calculus by Gilbert

Topic Beta & Gamma functions Statement of Dirichlet’s Theorem $ \int \int \int_{V} x^{l-1} y^{m-1} z^{n-1} dx dy ,dz = \frac { \Gamma {(l)} \Gamma {(m)} \Gamma {(n)} }{ \Gamma{(l+m+n+1)} } $ , where V is the region given by $ x \ge 0 y \ge 0 z \ge 0

We all know that the derivative of $x^2$ is 2x. But what if someone proves it to be just x?

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