Here is my assignment in all its entirety:
JTG 2012 Midterm
March 6, 2012
Due Friday, March 23, 5PM
Sometimes students will turn in this sloppy ____ in Journey, the kind of thing they would never turn in for a paper.
-Frank Kelly
Remember, the main reason for writing these things out is to practice your skills and philosophy of exposition, not simply display the correct answer. Be neat. Think like a designer, where justification is required and where it is unnecessary. How do you know what you know and is it reasonable to suppose your reader knows that too? Write with the aim of satisfying the curiosity of a fellow student, not proving to the teacher that you know the right answer. Your challenge is to walk the tightrope between pedantry and hurried arrogance.
If your handwriting is anywhere near messy, you would do better to type out what you can. Do NOT write on both sides of a page. It’s encouraged to work together and use the internet, but cite your sources of inspiration and make sure what you hand in can honestly be called yours.
Hints
- State the problem you are trying to solve.
- Use at least a few words.
- If you are asked to find something, you want to substantiate your claim that you’ve found the right thing.
- Read these instructions again when it matters.
Questions
1. Consider the sequence
2+1
2 x 3+1
2 x 3 x 5+1
2 x 3 x 5 x 7+1
2 x 3 x 5 x 7 x 11+1
generated by adding the next largest prime to the sequential product at each step. Find the first member of this sequence that is not a prime and factor it completely.
Bonus - Find a method for generating odd numbers that continues for more steps before giving a non-prime number.
2. Find all the factors of 496 and prove that 496 is a perfect number.
Bonus - Do the same for 8128 and 8,589,869,056. This will make you current with the best in 1588.3. Given a circle of radius one, find the length of a side of an inscribed, regular 24-gon by using the length of a side of an inscribed regular 12-gon (like we did in class). Use this length to approximate π by providing a fraction (square roots are okay, but decimals are not) that is an under-estimate of π.
Bonus - Go all the way to 96. This makes you current with Archimedes.4. Generate a cubic:
ax^3+bx^2+cx+d
where aand b are nonzero integers so that your cubic is the un-depressed version of the cubic in Chapter 6:
x^3+6x-20 .
Next, use the solution to the depressed cubic in Chapter 6 to find a solution to your cubic.5. Use Egyptian Multiplication to compute 57173. Then express your answer in binary notation.
Bonus - Now divide 571 by 73.6. Write a proof of the result that an angle inscribed in a semi-circle is a right angle. You may use the result that the sum of the angles of a triangle is 180º.
Bonus - Draw a diagram indicating where in Euclid’s Elements this occurs and what propositions are needed for its proof.7. Use Geometric Algebra to show that a^2-b^2=(a+b)(a-b). You can use the distributive rule in your demonstration. Do NOT use negative numbers.
8. In class we showed that 2 is an irrational number; that is, there is no fraction ab, where aand b are whole numbers and ab=2.
We did this by contradiction, assuming that such a fraction existed, noting that without loss of generality that that fraction could be written in lowest terms so that aand b are co-prime (they share no factors). In particular, this means that aand b CANNOT both be even. This gives us exactly 3 possibilities for the parity of aand b. If we rewrite ab=2 as a=b2, and if we look carefully, we can arrive at a contradiction in each of the three cases.
For this problem, I’m going to assume that you’ve done that, and know for a rock-solid fact that 2 is irrational. Given that fact:a. Show that 7+2 is an irrational number.
b. Show that 7+2 is an algebraic number. To do b., some additional background may be helpful.
Background for b.
We have seen rational numbers and irrational numbers. Rational numbers come from arithmetic of whole numbers, and our first source of irrational numbers is geometry. All the numbers that we can make with a compass and straightedge are called constructable numbers. In more modern times, we have another source of numbers: polynomial equations with rational numbers for their coefficients. Numbers who are solutions to these equations are called algebraic numbers.
Here are a couple of examples:
- 2is an algebraic number because it's a solution to the equation x2-2=0. If you plug 2 in for x, the equation is true.
- 5 is the solution to x-5=0. So it too is an algebraic number. Duh.
- Even though 2 is algebraic, it's not because x-2=0. To show that a number is algebraic, you can't use irrational numbers in the equation.
It's really easy to see that every rational number is algebraic. It's somewhat harder, but still reasonable to show that every constructable number is algebraic. What's amazing, difficult, and closely linked to the epilogue of chapter 6, is that not all algebraic numbers are constructable.
Bonus - Prove that 3 is irrational.
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