Enable contrast version

# Tutor profile: Alexander R.

Inactive
Alexander R.
Bachelor and Master's in Mathematics
Tutor Satisfaction Guarantee

## Questions

### Subject:Linear Algebra

TutorMe
Question:

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

Inactive
Alexander R.

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.

### Subject:Number Theory

TutorMe
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.

Inactive
Alexander R.

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.

### Subject:Discrete Math

TutorMe
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.

Inactive
Alexander R.

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 $$0.5073$$.

## Contact tutor

Send a message explaining your
needs and Alexander will reply soon.
Contact Alexander

Start Lesson

## FAQs

What is a lesson?
A lesson is virtual lesson space on our platform where you and a tutor can communicate. You'll have the option to communicate using video/audio as well as text chat. You can also upload documents, edit papers in real time and use our cutting-edge virtual whiteboard.
How do I begin a lesson?
If the tutor is currently online, you can click the "Start Lesson" button above. If they are offline, you can always send them a message to schedule a lesson.
Who are TutorMe tutors?
Many of our tutors are current college students or recent graduates of top-tier universities like MIT, Harvard and USC. TutorMe has thousands of top-quality tutors available to work with you.
BEST IN CLASS SINCE 2015
TutorMe homepage