$75/hour

4.9
average from
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“**Best Trig Tutor**”

Hi, my name is Rob. I graduated from SUNY Oswego summa cum laude with a BS in applied mathematics and a minor in statistics. I'm currently working as an actuarial consultant.

I was trained as a math tutor in 2010 and have since tutored privately, at Cayuga Community College, and at SUNY Oswego. I have also worked as a teaching assistant

Background Check:
Passed

In-person lessons

Because of Rob's patience he was able to prep

my daughter in just 4 weeks for the Trig Regents.

She was able to pass this time. Would recommend Rob!!!!!

Robert saved me with my online statistics class. I wasn't making sense of it on my own, but he had a way of explaining things that I could actually understand what was going on. Robert was so patient and helpful, it really made up for the lack of a classroom. I'm lucky I found him in time for my final! Highly recommended!!!

Robert just finished tutoring my son in calculus. As soon as we contacted Robert he set up a meeting location and a time. He was always punctual, professional, and easy to reach. Best of all, he is very personable; my son really likes him. Robert is very knowledgeable about the material and seems to enjoy it, so that my son enjoyed learning it. To quote my son "That hour goes by SO fast!" I will definitely use Robert again if I need to and will recommend him to my friends.

Math:

ACT Math, Science:

Biostatistics
Test Preparation:

ACT Math, GRE, SAT Math
Computer:

MATLAB
Elementary Education:

Elementary Math
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GRE
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As a mathematics major, I am well familiar with the topics covered in discrete mathematics. This class is the gateway to the higher math classes. The topics introduced here are revisited over and over as one works with mathematics, and are crucial to understand.

I am well familiar with the topics covered in this class, including laws of probability, combinations, permutations, subsets, power sets, relations, and types of functions.

I'm also prepared to assist in learning the proofs necessary for this course, including but not limited to induction, contra positive, contradiction, and if-and-only-if type proofs.

The following is a proof done by me on antisymmetric relations, followed by a shorter proof on set intersection of set differences. Set notation is displayed as "?" on this site, obscuring details

A relation R on a set A is antisymmetric if and only if R n R-1 ? {(a, a): a ? A}.

It will first be shown that if R n R-1 ? {(a, a): a ? A}, R on A is antisymmetric. Suppose R n R-1 ? {(a, a): a ? A}. Let R = {(x, y): x, y ? A}. By definition of inverse relation, R-1 = {(y, x): x, y ? A}. By definition of subset, if (x, y) ? R n R-1, (x, y) ? {(a, a): a ? A}. Hence, x = y, and so (x, y) ? R implies (y, x) ? R. Since x = y, ? (xRy ? yRx), x = y. Thus, by definition of antisymmetric, R on A is antisymmetric. Therefore, if R n R-1 ? {(a, a): a ? A}, R on A is antisymmetric.

It will now be shown that if R on A is antisymmetric, R n R-1 ? {(a, a): a ? A}. Let m, n ? A. Suppose that R on A is antisymmetric. Let R = {(m, n): m, n ? A}. Therefore, by definition of inverse relation, R-1 = {(n, m): m, n ? A}. Let (m, n) ? (R n R-1). By definition of intersection, (m, n) ? R and (m, n) ? R-1. Because (m, n) ? R-1, (n, m) ? R. Therefore, (m, n) ? R and (n, m) ? R. Since by definition of antisymmetric, (mRn ? nRm) implies m = n, then m = n. Notice that m and n are equal elements of A, and that (m, n) ? (R n R-1). Therefore, ? (m, n) ? (R n R-1), (m, n) ? {(a, a): a ? A}. Thus, by definition of subset, (R n R-1) ? {(a, a): a ? A}. Therefore, if R on A is antisymmetric, R n R-1 ? {(a, a): a ? A}.

It has been shown that if R n R-1 ? {(a, a): a ? A}, R on A is antisymmetric. It has also been shown that if R on A is antisymmetric, R n R-1 ? {(a, a): a ? A}. Therefore, a relation R on a set A is antisymmetric if and only if R n R-1 ? {(a, a): a ? A}.

Given that A and B are sets, (A - B) n (B - A) = { }.

Let A and B be sets. Suppose, for the sake of contradiction, (A - B) n (B - A) ? { }. Let x ? (A - B). By the definition of symmetric difference, x ? A and x ? B. Since (A - B) n (B - A) ? { }, by definition of intersection, ? at least one arbitrary element, x, in (A - B) and (B - A). By definition of symmetric difference, if x ? (B - A), x ? B and x ? A. ==><==.

Observe that if (A - B) n (B - A) ? { }, x ? A and x ? A. Because this cannot be true, it is not true that (A - B) n (B - A) ? { }. Therefore, (A - B) n (B - A) = { }.

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