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# Tutor profile: Michael G.

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Michael G.
Math major at Princeton
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## Questions

### Subject:Set Theory

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Question:

Say there are 2 wizards, each given the same infinite countable sequence of boxes, each containing a real number. Neither wizard can see the other, but they can open as many of their own boxes as they wish. They may also deliberate for an infinite amount of time before receiving the boxes, and may open boxes for an infinite amount of time. At the end of the game, each of them will point to an unopened box, and claim what number they think is in the box. Find a strategy such that at least one of them will be right. Assume the axiom of choice.

Inactive
Michael G.

To understand this problem, it is important to understand the axiom of choice. The axiom of choice states that from any set of sets, regardless of the sizes of the sets, a set can be constructed containing exactly one element from every set. Sometimes the axiom of choice isn't necessary; for example, given a set of sets each containing a left shoe and a right shoe, we can simply define a rule to pick an element, such as always picking the left shoe (or if they are numbers, the smaller number). However, in this case we will end up needing it. Since we know we want to use the axiom of choice, we want to find a set of sets in which it would be nice to pick an element from each set. Since we are looking at sequences of numbers, we may want this set of sets to be a partition of all possible sequences of real numbers. Here is the tricky part: we will put two elements in the same set when they differ in a finite number of places. For example if two sequences differ in the first position, the tenth position and the ninetieth position, but are the same in every other position, we will put them in the same set. By the axiom of choice, for any set $$S$$ we can pick an element $$A(S)$$ of that set to represent it. Now let's take a look at the wizard's strategy. Let's have the first wizard open the boxes in the even numbered positions, and have the second wizard open the boxes in the odd numbered positions. Now each of them has opened an infinite sequence of boxes. Each of these sequences $$S$$ has a "representative sequence" $$A(S)$$ which it differs from in finitely many places. The key to this proof is that one of these sequences has to have the furthest position where it differs from its representative. Each of them should keep closed the box just to the right of the furthest one that they open which differs from that sequence's representative, and open all of the other previously closed boxes. Now the person who originally opened the even boxes has all of the odd boxes opened but one. This sequence of odd boxes has a representative, because even though one box is closed, the sequence will differ in finitely many places regardless of if the representative differs from the closed box. The wizard should guess the value that the representative thinks is in that box. The other wizard should do something similar. Now if originally the furthest difference from the representatives was in an odd position, then the wizard guessing an even box will be right, and vice versa.

### Subject:Discrete Math

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Question:

Consider a point A on the corner of a $$n$$ dimensional cube with side length 1. A corner B of the cube is chosen at random (B could be the point A itself). What is the expected value of the square of the distance from A to B?

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Michael G.

The expected value of a random variable X with a finite set S of possible values can be expressed as $(\sum_{x\in S} (x\cdot P(X=x)).$) This is to say that if we multiply the probability of each value occurring by the value itself, and add all of these products together, we will arrive at the answer. Luckily for us, these values are not as hard to calculate as they seem. This cube has $$n$$ dimensions, each of which has 2 sides that a corner could be on (for example, in 3 dimensions, a point could be on the top side or the bottom side). Say that B is on the other side as A in $$k$$ dimensions. Then to get from A to B, we must move one unit in $$k$$ perpendicular directions. Then by the Pythagorean Theorem, the distance between the two points is $$\sqrt{k}$$. Since we are looking at the square of the distances, we now know that these squares must be integers, which gives us a lot more freedom. We can now try to consider the relationships between points. Specifically, consider points that are on "opposite" corners of the cube (meaning they are on the other side as each other in every dimension). If one of them differs in $$k$$ dimensions, then the other differs in $$n-k$$ dimensions. Therefore we can split the points into pairs such that the average value of the square of the distances from A is $$n/2$$. Therefore the sum of the squares of the distances of all the points is $$2^{n}\cdot n/2$$, so by our expected value formula, our answer is $$n/2$$.

### Subject:Algebra

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Question:

Prove that for all positive $$a,b,c$$ such that $$a+b+c=1$$, $(27(ab+bc+ca)\leq 1/a+1/b+1/c$)

Inactive
Michael G.

Note that we can divide the right hand side by $$abc$$, turning the equation to $(27abc(1/a+1/b+1/c)\leq1/a+1/b+1/c$) We can divide both sides by $$(1/a+1/b+1/c)$$: $(27abc\leq1$) Now the AM-GM inequality states that the arithmetic mean of a set is always less than or equal to that set's geometric mean. This is convenient for us, since we know the arithmetic mean of $a,b,c$ is 1/3. So the inequality states $(1/3\geq\sqrt[3]{abc}$) $(1/27\geq abc,$) and this is the desired inequality.

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