Mar 092016
 

This is going to be another one of my “selfish” posts – written primarily for me to refer back to in the future and not because I believe it will benefit anyone other than me. The idea is one that I always took for granted but had a hard time proving to myself once I decided to try.

Theorem: Suppose we have an M bit unsigned binary integer with value A. Consider the first (least significant) N bits with value B. Then:

A \equiv B \bmod{2^N}

Put another way, arithmetic with unsigned binary integers of a fixed length N is always performed modulo 2^N.
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Jan 122016
 

It is no big secret that exponentiation is just multiplication in disguise. It is a short hand way to write an integer times itself multiple times and is especially space saving the larger the exponent becomes. In the same vein, a serious problem with calculating numbers raised to exponents is that they very quickly become extremely large as the exponent increases in value. The following rule provides a great computational advantage when doing modular exponentiation.

The rule for doing exponentiation in modular arithmetic is:

a^b \bmod{c} = ((a \bmod{c})^b) \bmod{c}

This states that if we take an integer a, raise it to an integer power b and calculate the result modulo c we will get the same result as if we had taken a modulo c first, raise it to b, and calculate that product modulo c.

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Jan 032016
 
The Complete Idiot's Guide to Calculus Book Cover The Complete Idiot's Guide to Calculus
W. Michael Kelley
Mathematics
Penguin
2006
336

Let's face it: most students don't take calculus because they find it intellectually stimulating. It's not . . . at least for those who come up on the wrong side of the bell curve! There they are, minding their own business, working toward some non-science related degree, when . . . BLAM! They get next semester's course schedule in the mail, and first on the list is the mother of all loathed college courses . . . CALCULUS! Not to fear-The Complete Idiot's Guide to Calculus, Second Edition, like its predecessor, is a curriculum-based companion book created with this audience in mind. This new edition continues the tradition of taking the sting out of calculus by adding more explanatory graphs and illustrations and doubling the number of practice problems! By the time readers are finished, they will have a solid understanding (maybe even a newfound appreciation) for this useful form of math. And with any luck, they may even be able to make sense of their textbooks and teachers.

Jan 022016
 

I must stay focussed. I must stay focussed. I must stay … I wonder what’s new on Facebook.

I don’t really feel like writing this post mostly because I know that it will be very similar to the other two I have already done: modular addition rule proof and modular subtraction rule proof, but my New Year’s Resolution is to follow things through to completion. Well, that would’ve been my News Years resolution if I had made one. Either way, it’s back to modular arithmetic.

The rule for doing multiplication in modular arithmetic is:

(a * b) \bmod{c} = (a \bmod{c} * b \bmod{c}) \bmod{c}

This says that if we multiply integer a times integer b and take the product modulo c, we get the same answer as if we had first taken a modulo c and multiplied by b modulo c and taken that product modulo c.

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Dec 312015
 

I’ve already presented and proved the rule for modular addition, so for a sense of completeness, but mostly to satisfy my OCD, now I’ll cover the rule for modular subtraction. When doing subtraction in modular arithmetic, the rule is:

(a - b) \bmod{c} = (a \bmod{c} - b \bmod{c}) \bmod{c}

If we subtract integer b from integer a and calculate the difference modulo c, we get the same answer as if we had subtracted b modulo c from a modulo c and then calculated that difference modulo c. Like the modular addition rule, this rule can also be expanded to include multiple integers. Continue reading »

Dec 312015
 

Addition in modular arithmetic is much simpler than it would first appear thanks to the following rule:

(a + b) \bmod{c} = (a \bmod{c} + b \bmod{c}) \bmod{c}

This says that if we are adding two integers a and b and then calculating their sum modulo c, the answer is the same as if we added a modulo c to b modulo c and then calculated that sum modulo c. Note that this equation can be extended to include more than just two terms. Continue reading »

Nov 152014
 
Leonhard Euler

Leonhard Euler

Besides being an obvious lady killer, Swiss mathematician Leonhard Euler gifted the world with some pretty important mathematical concepts, notational conventions, and formulas. I almost feel bad about the fact that I couldn’t even spell his name correctly until I was well into adulthood.

You are probably thinking, “Sure, he had a bitchin’ robe, and for an old-timey dude he was pretty good looking, but was he really all that?”

I can’t find any quotes from the ladies, but his peers, both contemporary and modern have had a few things to say about him and his work.

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Nov 102014
 

(Note: I lied. This will be my first “neural dump.” I began writing about Euler’s Formula, but felt what follows was worthy of its own post and a better foundation for what will follow when I tackle Euler.)

Complex numbers arose from the fact that there is no solution for x in the equation x^2 = -1 in \mathbb{R}, the set of real numbers.

Early mathematicians being the devil-may-care mavericks that they were, were all like, “Screw it. Let’s just invent a new number. We’ll just call this number i and say that the solution is x = i.” Or, in other words, this new number i they imagined up (see what I did there) is equal to \sqrt{-1}.

    \[i = \sqrt{-1}\]

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