# Tutor profile: Tristan V.

## Questions

### Subject: Set Theory

The symmetric difference $$A\mathrel\vartriangle B$$ is defined as the set $(A\mathrel\vartriangle B=(A\setminus B)\cup(B\setminus A).$) Show that $((A\mathrel\vartriangle B)\mathrel\vartriangle C=A\mathrel\vartriangle (B\mathrel\vartriangle C)$)

1. ---- Let $$x\in(A\mathrel\vartriangle B)\mathrel\vartriangle C$$, then $$x\in(A\mathrel\vartriangle B)\setminus C$$ or $$x\in C\setminus(A\mathrel\vartriangle B)$$. In the first case we see that both $$x\notin C$$ and $$x\in(A\mathrel\vartriangle B)$$, so $$x\in A\setminus B$$ or $$x\in B\setminus A$$. Therefore we have [$$x\in A$$ and $$x\notin B$$ and $$x\notin C$$], or we have [$$x\notin A$$ and $$x\in B$$ and $$x\notin C$$]. In the second case we see that both $$x\in C$$ and $$x\notin (A\mathrel\vartriangle B)$$. Because $$x\notin(A\mathrel\vartriangle B)$$ means that not [$$x\in A\setminus B$$ or $$x\in B\setminus A$$], we see (by De Morgan's law) that $$x\notin A\setminus B$$ and $$x\notin B\setminus A$$. Therefore either [$$x\notin A$$ and $$x\notin B$$ and $$x\in C$$], or [$$x\in A$$ and $$x\in B$$ and $$x\in C$$]. So we see that $$x\in(A\mathrel\vartriangle B)\mathrel\vartriangle C$$ if and only if $$x$$ is contained in all three sets, or $$x$$ is contained in exactly one of the three sets. We can do the steps as we did above, to show that the same is true for $$x\in A\mathrel\vartriangle(B\mathrel\vartriangle C)$$. It follows that $$(A \mathrel\vartriangle B)\mathrel\vartriangle C = A\mathrel\vartriangle ( B\mathrel\vartriangle C)$$.

### Subject: Discrete Math

Prove by induction that $$n!>2^n$$ for all $$n\geq 4$$.

For the base case, let $$n=4$$, then we see that $$n!=2\cdot3\cdot4=24>16=2^4$$, so the statement that $$n!>2^n$$ is true for $$n=4$$. For the induction case, assume as induction hypothesis that $$n!>2^n$$ is true for some $$n\geq 4$$. We have to show that $$(n+1)!>2^{n+1}$$. Since $$(n+1)!=(n+1)\cdot n!$$, we know by the induction hypothesis that $$(n+1)\cdot n!>(n+1)\cdot 2^{n}$$. But because $$n\geq 4$$, we know that $$(n+1)\cdot 2^{n}>2\cdot 2^n=2^{n+1}$$. Therefore $$(n+1)!>2^{n+1}$$.

### Subject: Algebra

What are the solutions to the following equations: 1. $$2x+3=0$$ (early high school) 2. $$2x^2+3x-5=0$$ (late high school) 3. $$2x^4 + 3x^3 + 18x^2 + 27x=0$$ (uni calculus)

1. ---- We want to isolate $$x$$ on one side of the equation and have only numbers on the other side. So first, let's subtract $$3$$ from both sides of the equation: $( 2x+3=0\\ \Downarrow \\ 2x=3 $) Next, let's divide by $$2$$ on both sides to get the answer: $( 2x=3\\ \Downarrow \\ x=\frac{3}{2} $) 2. ---- First, let's divide the expression by $$2$$ to get rid of the factor in front of $$x^2$$: $( 2x^2+3x-5=0\\ \Downarrow \\ x^2+\tfrac32x-\tfrac52=0 $) We will try to factor this quadratic polynomial to get an expression of the form $$(x+a)(x+b)$$. We need to find $$a$$ and $$b$$ such that $$a\cdot b=-\tfrac52$$ and $$a+b=\tfrac32$$. With some searching we can find that $$a=-1$$ and $$b=\tfrac52$$ work: $( x^2+\tfrac32x-\tfrac52=(x-1)\left(x+\tfrac52\right) $) Therefore the solution is $$x=1$$ or $$x=-\tfrac52$$. 3. ---- We try to factor $$2x^4 + 3x^3 + 18x^2 + 27x$$. Note that $$x$$ is a common factor, thus $(2x^4 + 3x^3 + 18x^2 + 27x=2x\cdot(x^3 + \tfrac32x^2 + 9x + \tfrac{27}2)$) Therefore $$x=0$$ is one solution. We will factor $$x^3 + \tfrac32x^2 + 9x + \tfrac{27}2$$ by splitting it in two parts: $( x^3 + \tfrac32x^2 + 9x + \tfrac{27}2 = (x^3 + \tfrac32x^2) + (9x + \tfrac{27}2)\\ \qquad\qquad\qquad\quad\ = x^2(x + \tfrac32) + 9(x + \tfrac{3}2)\\ \qquad\qquad\quad= (x^2+9)(x + \tfrac32) $) Therefore $$x=-\frac32$$ is another solution. Finally we solve $$x^2+9=0$$: $( x^2=-9\\ x=\pm\sqrt{-9}\\ x=\pm 3i $) Thus the two real roots are $$x=0$$ and $$x=-\frac32$$, and the two complex roots are $$x=3i$$ and $$x=-3i$$

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