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Alexander R.

Bachelor and Master's in Mathematics

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Linear Algebra

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Question:

If the nullity of a linear map is zero, then independent vectors are mapped to independent vectors.

Alexander R.

Answer:

Let $$\{u_1, \cdot\cdot\cdot u_n\}$$ be a set of linearly independent vectors and $$A$$ be a linear map such that $$Null(A) =\{0\}$$. Then consider the $$v = \Sigma_{i=1}^{n}c_i u_i$$ where $$A(v) = 0$$. This implies that $$v\in Null(A)$$ and therefore $$ $v = \Sigma_{i=1}^{n}c_i u_i = 0$$. Since $$\{u_1, \cdot\cdot\cdot u_n\}$$ are linearly independent then $$c_i=0$$ for all $$i$$. So $$A(v) = A(\Sigma_{i=1}^{n}c_i u_i) = \Sigma_{i=1}^{n} c_i A(u_i) = 0$$ and that $$c_i=0$$ for all $$i$$. Therefore $$\{A(u_1), ...,A(u_n)\}$$ are linearly independent.

Number Theory

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Question:

Let n be a positive number. If the sum of the digits in n are divisible by 9 then n is divisible by 9.

Alexander R.

Answer:

If n is a number then it can be written as followed $$n = \Sigma_{i=0}^{n}d_i\cdot 10^i$$, where $$d_i$$ is a number between $$[0,9]$$. From the rules of operation for modular arithmetic $$ab\ \text{mod(r)} = (a\ \text{mod(r)})(b\ \text{mod(r)})\ \text{mod(r)}$$, therefore given $$10\ \text{mod(9)} = 1$$, then $$10^i \ \text{mod(9)} = 1$$. If $$\Sigma_{i=0}^{n}d_i = 9k$$ for some k and $$n = \Sigma_{i=0}^{n}d_i\cdot 10^i$$, then $$n\ \text{mod(9)} = \Sigma_{i=0}^{n}d_i\cdot 10^i\ \text{mod(9)} = \Sigma_{i=0}^{n}d_i\cdot 1\ \text{mod(9)} = 9k\ \text{mod(9)} = 0$$. Therefore n is divisible by 9.

Discrete Math

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Question:

The birthday problem. How many people need to be in a room such that the probability that two of them share a birthday is greater then or equal to 50%? Assume birthdays are uniformly distributed.

Alexander R.

Answer:

The first thing to realize about this problem is that it is easier to determine its compliment. If we want to know if the probability two people share a birthday then the compliment is the probability that two have unique birthdays. Let $$A_i$$, represent the event that person $$i$$ in the room has a unique birthday from the other $$i-1$$ people in the room. Lets count probability that r people don't share a birthday. If $$r = 2$$, then the first person has probability of 1 of having a unique birthday $$P(A_1) = 1$$ and the second person has 364 days out of 365 possible days to be their birthday unique from person 1, $$P(A_2) = \frac{364}{365}$$. Given peoples birthdays are independent $$P(A_1\cap A_2) = P(A_1)P(A_2) = \frac{364}{365}$$. If $$r=k$$, then $$P(A_1) = 1$$, $$P(A_2) = \frac{364}{365}$$, ... $$P(A_k)=\frac{365 - k+1}{365}$$. $$P(A_1\cap A_2\cap \cdot\cdot\cdot \cap A_k) = \frac{365}{365}\cdot \frac{364}{365}\cdot\cdot\cdot \frac{365 - k+1}{365}$$. If the probability that k people in the room have unique birthdays. Therefore $$1-P(A_1\cap A_2\cap \cdot\cdot\cdot \cap A_k)$$, represents the probability that two people in a room with k people share a birthday. Using the formula above it can be shown that when $$r=23$$ the probability that two people share a birthday is approximately $$0.5073$$.

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