Contemporary Topics in Math: Test 1

Absolute Value Algorithm

Input: An integer a Output: An integer |a| Case I: If a ≥ 0:Return |a| = a Case II: If a < 0: Return |a| = −a

Naive Exponentiation

Input: An integer b and a natural number n Output: cn = b^n

Division algorithm puts out this theorem, but we can also right the notation in other ways

Let a be an integer and b be a natural number. Then, there exist unique integers q and r with 0 ≤ r < b such that a = b·q + r. Other notations: Let a be an integer and b be a natural number, and let q and r be the unique integers such that 0 ≤ r < b and a = b·q + r. We write: a÷b = q remr Let a be an integer and b be a natural number. We denote the remainder of the division of a by b by: a mod b = r

Beizout's Thm

Let r0 and r1 be natural numbers. Then, there exist integers s and t with sr0 + tr1 = gcd(r0,r1)

Integers

Numbers (...-1, 0, 1...)

Additive Communative

a + b = b + a

Additive Closure

a + b is an integer

Negative Integers

... -3, -2, -1

Additive Identity

0 + a = a

Multiplicative Identity

1 x a = a

Positive Integers/Natural Numbers

1, 2, 3, ...

Multiplicative Communative

a x b = b x a

Multiplicative Closure

a x b is an integer

Pythagorean Triples

three integers a, b, and c that satisfy the equation a2 + b2 = c2. We write Pythagorean triples in the form (a,b,c) but have to be written from smallest to largest.

Property of Exponents

B^n x B^m = Bm+n (B^m)^n = Bmxn (axb)^n = a^n x b^n

Additive Associative

(a+b) +c = a + (b+c)

Multiplicative Associative

(axb) x c = a x (bxc)

Multiplicative Inverse

-1 x a = -a

Make sure you also know:

-ISBN - how to find if numbers are valid, or find a missing # (mod 11!!!!), It is valid when you get a remainder of 0 -Find LCM, GCD -Beizout's Thm - finding S and T

Algorithm

A finite sequence of instructions performing a task.

Additive Inverse

For each non-zero a there is a unique non zero integer -a such that: a + (-a) = 0 (-a is the additive inverse)

Pythagos Proof (Pythagorean thm)

Picture one takes out a2 and b2 (Take out two perfect squares and you are left with the two rectangles), now for picture two you have the perfect square with the second perfect square inside of it - slightly tilted, the square in the middle is side c giving you c2.

Bhaskara Proof (Pythagorean thm)

The top picture is a square with side lengths c and overall area of c2. It is tiled by four copies of a right triangle with legs of lengths a and b and hypotenuse of length c, and a square with side lengths b−a. Rearrange the five shapes used to build the top picture to create the picture on the bottom left. Overall, it is a square with side lengths a sitting next to a square with side lengths b, having a total area of a2 + b2.

Fermat's Last Theorem

There are no natural numbers a, b, and c that satisfy the equation an + bn = cn for any integer n > 2.