# 76 Youtube Calculus Tutorials and More by Prof Leonard

For those who are studying undergraduate calculus, Prof Leonard is another addition to the video tutorials I have already shared in Math and Multimedia. The Prof Leonard channel contains 76 Calculus I, II, and III videos ranging from 15 minutes up to more than 3 hours. Most of the videos are about 1 hour in length. The channel also contains videos on Intermediate Algebra and Statistics.

I have shared several Youtube channels in this blog about calculus particularly that of Khan Academy, MIT Open Courseware, and Patrick JMT tutorials. You can also visit more video tutorials here.

# Brilliant: A Problem Solving and Competition App

Last September, I have posted about MathCounts Trainer, a math competition app for iOS devices. In this post, I am going to introduce to you Brilliant, another math competition app which is both available in iOS and Android devices.

Brilliant is an app created by Brilliant.org intended to train users for problem solving, test preparation, and mathematics and science competitions. It has a wide range of problems in terms of difficulty ranging from basic concepts to Olympiad level. It includes problems in Basic Math, Algebra, Geometry, Calculus, SAT, JEE, Physics, Computer Science, and Quantitative Finance.

In Brilliant, you can select the subject that you want to learn and choose the topic or level you want to study. It contains thousands of problems created by a community of mathematicians and scientists. Users of this app can participate in discussions, follow other members, and learn from the community.

# The Algebra and Geometry of Square Root

If we have a square with a given area, then we can find the length of its side. For example, a square with area 4 square units has a side length of 2 units. In other words, in finding the side length of a square with area 4 square units,we are looking for a number that is equal to 4 when squared.The number that when squared is equal to 4 is called the square root of 4 and is written as $\sqrt{4}$. From the discussion above, we now know that

$\sqrt{4} = 2$.

It is easy to see that $\sqrt{1} = 1$, since $1^2 = 1$ and $\sqrt{0} = 0$ since $0^2 = 0$.

The square root of the two numbers above are integers, but this is not always the case. For instance, $\sqrt{2}$ is clearly not an integer since $1^2 = 1$and $2^2 = 4$. This means that $\sqrt{2}$ is somewhere between 1 and 4. What about $\sqrt{5}$Continue reading