You might have noticed that in this blog, I am very persistent in providing proofs for anything I say. This is because I remember wondering in math class why everything was true, and I think that information should be provided.
Most mathematicians are the same way, where they value proofs of theorems, and will never accept something just cause it seems to work a couple times. In fact, one of the main jobs of a mathematician is to take unproved conjectures and prove them.
One of these unproven conjectures is the Riemann Hypothesis, which I have talked about on this blog before. This is a problem where mathematicians have been working for decades to find a proof.
Godfrey Hardy was a British mathematician who valued proofs. He was born February 7, 1877, and had a huge contribution to the fields of number theory and mathematical analysis. He once said to a colleague, "If I could prove by logic that you would die in five minutes, I should be sorry you were going to die, but my sorrow would be very much mitigated by pleasure in the proof."
You might recognize Hardy as the mentor of Srinivasa Ramanujan, who I posted about a few months ago. When Hardy was asked what his biggest contribution to mathematics was, he stated without hesitation that it was his discovery of Ramanujan.
Hardy was an atheist, but liked to play games with a Godlike being, which him and other mathematicians like to call the Supreme Fascist, or SF. One of these stories was one that I enjoyed, and wanted to share.
When he was on a boat ride from Scandinavia home to England, the water started to get very turbulent. So, he sent a postcard to a colleague back home saying that he had proved the Riemann Hypothesis. Knowing the importance of the problem, he didn't think the SF would let him die with everyone thinking he had a proof, so he felt that his safety was then guaranteed.
Saturday, July 27, 2013
Saturday, July 20, 2013
Multiplication Made Visual
In school, multiplication is taught by giving students a specific method to use (which they call the "traditional" method), and you have to solve problems with it. By only providing one of over a dozen multiplication methods, this eliminates the creativity, and quite possibly the fun, in multiplication. I'm not going to dwell on these issues; you can read my Capstone Research Paper for that.
I have discussed a few other methods of multiplication on this blog, one of which is the CrissCross Method. Again, I won't explain it here, but this method does lay the foundation for the new method I am going to explain today.
Rather than taking you through the steps, I will post a video tutorial on how to do it (those are usually more fun anyways). If you are wondering why it works, I encourage you to refer back to the CrissCross Method and notice the similarities. Try a few problems with the CrissCross Method, and then this method, and see how the numbers that appear throughout the problem seem to be identical.
When you watch this video, I would strongly suggest doing the examples along with the video. It can be a very useful technique for problems whose digits are of a reasonable size.
I think this is a really cool method of multiplication, and even easier to teach than the traditional method. Play around with it, and you will likely end up using it in the future.
For some more intellectual readers, it might be fun to try multiplying numbers with bigger digits using the quinary or senary number systems (base five or base six). This would make less dots in each section. I think it might be a fun exercise.
Answers to June Problem of the Week:
*For the answers to the creation of a line of best fit, it will be written as an inequality. There is no way to get the exact answer by hand, so just make sure your answer is in the interval that is written.
Easy:
a = 7^{1}/_{3}
p = 14
I have discussed a few other methods of multiplication on this blog, one of which is the CrissCross Method. Again, I won't explain it here, but this method does lay the foundation for the new method I am going to explain today.
Rather than taking you through the steps, I will post a video tutorial on how to do it (those are usually more fun anyways). If you are wondering why it works, I encourage you to refer back to the CrissCross Method and notice the similarities. Try a few problems with the CrissCross Method, and then this method, and see how the numbers that appear throughout the problem seem to be identical.
When you watch this video, I would strongly suggest doing the examples along with the video. It can be a very useful technique for problems whose digits are of a reasonable size.
I think this is a really cool method of multiplication, and even easier to teach than the traditional method. Play around with it, and you will likely end up using it in the future.
For some more intellectual readers, it might be fun to try multiplying numbers with bigger digits using the quinary or senary number systems (base five or base six). This would make less dots in each section. I think it might be a fun exercise.
Answers to June Problem of the Week:
*For the answers to the creation of a line of best fit, it will be written as an inequality. There is no way to get the exact answer by hand, so just make sure your answer is in the interval that is written.
Easy:
a = 7^{1}/_{3}
p = 14
m = ^{118988}/_{6925}
n = ^{394}/_{277}
y = 15.76
z = 12.08
g = 8
b = 112
q = 44
x = 33
Medium:
h = 105
p = 270
t = 19
n = 92925
u = 92.925
q = 135
a = 2165.2875
d = 21.652875
g = 15
j = 46.4625
k = 23.23125
r = 27
s = 45
2 < m < 3.5
45 < b < 50
175 < x < 275
Hard:
s = 107
t = 365
p = 905
q = 4096
q_{0} = 4096
q_{1} = 2048
q_{2} = 1024
q_{3} = 512
q_{4} = 256
q_{5} = 128
q_{6} = 64
q_{7} = 32
q_{8} = 16
q_{9} = 8
q_{10} = 4
q_{11} = 2
q_{12} = 1
q_{13} = ^{1}/_{2}
q_{14} = ^{1}/_{4}
q_{15} = ^{1}/_{8}
x_{1} = 452.5
x_{2} = 2
x_{3} = 4
y = 0
a = 226.25
b = 224.25
c = 454.5
t = 2635773.77
m = 34
x = 79.53%
n = ^{394}/_{277}
y = 15.76
z = 12.08
g = 8
b = 112
q = 44
x = 33
Medium:
h = 105
p = 270
t = 19
n = 92925
u = 92.925
q = 135
a = 2165.2875
d = 21.652875
g = 15
j = 46.4625
k = 23.23125
r = 27
s = 45
2 < m < 3.5
45 < b < 50
175 < x < 275
Hard:
s = 107
t = 365
p = 905
q = 4096
q_{0} = 4096
q_{1} = 2048
q_{2} = 1024
q_{3} = 512
q_{4} = 256
q_{5} = 128
q_{6} = 64
q_{7} = 32
q_{8} = 16
q_{9} = 8
q_{10} = 4
q_{11} = 2
q_{12} = 1
q_{13} = ^{1}/_{2}
q_{14} = ^{1}/_{4}
q_{15} = ^{1}/_{8}
x_{1} = 452.5
x_{2} = 2
x_{3} = 4
y = 0
a = 226.25
b = 224.25
c = 454.5
t = 2635773.77
m = 34
x = 79.53%
Friday, July 19, 2013
Problem of the Week Day 5: Week of 7/15/13  7/19/13
Today is the final day of the problem of the week. In all three problems, your goal is to find the value of x. Good luck!
Easy:
Find the mean of the following list of numbers:
a
b
f
g
q
x = ____
Medium:
The data set below represents the test scores of nine students based on the amount of absences they had.
x

y

f  f_{1 } f_{2}  f_{3}  f_{4}

ceiling(n ÷ f  r)

floor(f_{4 } f_{2 } r)

floor(8r)

floor(f_{2 }÷ f_{3})

ceiling(2f_{4 } r)

^{3}√(f_{1 } f_{4})

f ÷ 2

ceiling(f_{2 } f_{3 } r)

2f_{2}

√(f_{3})

floor(f_{1 } r)

floor(n ÷ 1000)

floor(n ÷ 100)

f_{1 } f_{4}

f_{1}

f_{4 } f_{2}

f_{4}

Find the line of best fit for this data set. It should be of the form y = mx + b.
m = ____
b = ____
Now, find the predicted score of a student with ten absences.
x = ____
Hard:
Take the following matrix that represents the payoffs in a mathematical game.

A

B

1

b, w  6

s, t

2

b, a

a, c

3

g, d

f, b

After eliminating any dominated strategies, use mixed strategy Nash equilibria to determine the percent of the time that strategy 1 should be played.
x = ____
The answers to the problem will be up in a month. I will post the answers to June’s problem of the week with tomorrow's post.
Thursday, July 18, 2013
Problem of the Week Day 4: Week of 7/15/13  7/19/13
Today is day four of the problem of the week. Good luck!
Easy:
Take the following parallelogram.
Assume that this parallelogram has an area of j. Find the measure of the base b and the perimeter q.
b = ____
q = ____
Medium:
Take the following ellipse:
Assume that this ellipse has an area of n. Find the measure of the short radius r. Round to the nearest tenth.
r = ____
Hard:
Take the following two triangles:
Determine the length of w. Round to the nearest integer.
w = ____
Hint: Look for triangle congruence or proportionality.
Wednesday, July 17, 2013
Problem of the Week Day 3: Week of 7/15/13  7/19/13
Today is day three of the problem of the week. Before I begin the problems, I would like to explain one thing, which will be used throughout the medium and hard problems. You might not have learned it in school, but the concept is very simple.
A floor function is the largest integer less than or equal to a given number. For example, the floor function of 6.7 is 6. The floor function of 4.973 is 4. The floor function of Ï€ is 3. Basically, if you round the number down to the nearest whole, you will have its floor function.
Similarly, a ceiling function is the smallest integer greater than or equal to a given number. So, the ceiling function of 6.7 is 7. The ceiling function of 4.973 is 5. The ceiling function of Ï€ is 4. While rounding down yields the floor function, rounding up gives the ceiling function.
Normally, this is denoted by a special sort of bracket. Since I don’t know how to type these brackets into the computer, I will use the following notation:
floor(x) = the floor function of x
ceiling(x) = the ceiling function of x
This is an easy way to eliminate fractions and decimals from numbers to make the problems slightly easier and more realistic.
Good luck!
Easy:
What is the Least Common Multiple of f, g, and h? Use the letter j to denote the answer.
j = ____
Medium:
What is the number of dots in a regular (f_{4}  f_{2})gon array whose sides are of length (f_{1}  f_{3})?
n = ____
Hard:
Find the explicit formula for the following sequence:
v, t, s, floor((g + f)/10), (g  60)/2, d  2, ...
The formula should be of the form ax^{2} + bx + c. So, write your answer in terms of the value of coefficients a, b, and c.
a = ____
b = ____
c = ____
Tuesday, July 16, 2013
Problem of the Week Day 2: Week of 7/15/13  7/19/13
Today is day two of July’s problem of the week! For the medium and hard problems, today is when you will start to use variables. For the easy problem, plug in the variables from yesterday for today.
Easy:
Using the answers from yesterday, solve the following problems for f, g, and h. Remember to use the order of operations.
f = 2(a) + a ÷ 10  2
g = m^{2} + (m  2)(m  7)
h = p^{2} ÷ 18  (p ÷ 30)^{3}
f = ____
g = ____
h = ____
Medium:
The graphs of the four lines from yesterday should form a quadrilateral on your graph paper. Ignore the rest of the lines, and just analyze each segment forming the quadrilateral.
First, find the length of the each of the four segments, and use their names (ex: f_{1}) for the variables equal to them. So, f_{1} = the length of the segment that was originally the graph of f_{1}.
f_{1} = ____
f_{2} = ____
f_{3} = ____
f_{4} = ____
Then, find the perimeter of this quadrilateral. Call this perimeter f.
f = ____
Hard:
First, plot points P and Q on the graph’s xintercepts and point R on the graph’s positive yintercept. Then, find all six measurements within triangle PQR. Use d to denote the measure of angle P, f to denote the measure of angle Q, g to denote the measure of angle R, s to denote the measure of segment PQ, t to denote the measure of segment PR, and v to denote the measure of segment QR. Round all angle measures to the nearest degree.
d = ____
f = ____
g = ____
s = ____
t = ____
v = ____
Monday, July 15, 2013
Problem of the Week Day 1: Week of 7/15/13  7/19/13
Today begins the second of three 2013 problems of the week. If you remember from last month, there are three different problems: easy, medium, and hard. The easy problem is at a level where you will be fine with just a good understanding of foundational mathematics (around a sixth grade level). The medium problem requires an understanding of Algebra I, but nothing beyond that. The hard problem requires Geometry, Algebra II, and depending on the rigor of the classes, some content from a Trigonometry or Precalculus class might be needed as well. To find your level, try looking at the categories of my blog posts. Basic posts correspond well with the easy problem, intermediate posts correspond with the medium, and advanced correspond with the hard.
Each of the problems is broken into five parts, and I give each part on the next day of the week. On Friday, you will have to solve for x, which is the final answer. The answer to the problem gets posted after a month. So, remember to save your results from each day to plug into the next day’s problem.
Last month’s easy problem had a lot of heavy arithmetic in it, which probably made it complicated to solve. However, this problem will be much less strenuous, and focuses more on the procedures than the arithmetic.
Good luck!
Easy:
Take the following right triangle:
You are going to determine three things about this triangle. First, find the area, which will be denoted by the letter a.
a = ____
Now, find the perimeter, which will be p.
p = ____
Lastly, find the measure of the missing angle up top, which we’ll call m.
m = ____
Medium:
Normally I would start with some triangle computations, but this week, I chose to do some function graphing on Monday, and then use triangles and trigonometry on Tuesday.
Take out some graph paper and draw in your x and y axes. Then, graph the following four functions, who we’ll call f_{1}, f_{2}, f_{3}, and f_{4}.
f_{1}) y = 48  ^{4}/_{3}x
f_{2}) 10x  24y  360
f_{3}) 3(x  20) = 4y
f_{4}) 5(y  12) = 6(x + 15)
You won’t need to solve for any variables today, but keep the piece of paper that you graphed it on. This graph will be used in tomorrow’s problem.
Hard:
Similar to the medium problem, today will be focused on graphing and tomorrow will incorporate the trigonometry.
Take out a piece of graph paper (separate from the medium problem if you are doing both) and graph the following equation:
16y^{2} = 3(48  3x^{2})
Tomorrow’s problem will be using this graph, so don’t lose it.
Saturday, July 13, 2013
Math in the News: The Twin Prime Conjecture
A common question people wonder is "What do mathematicians do?" People know that scientists make advancements in science and engineers make advancements in engineering, but what about math? Similarly, mathematicians make advancements in math.
This is often an odd concept to people, since math seems like it is all figured out. How could it be taught so definitively if there are still things to figure out? But, there are tons of conjectures out there that people are trying to prove.
One of them is called the Twin Prime Conjecture. You are probably aware that a prime number is a number that is only divisible by one and itself. There are an infinite number of prime numbers (click here to see a proof). Twin primes are two prime numbers that differ by two. For instance, 3 and 5 are twin primes because they are both prime and differ by two.
We know that there are an infinite number of primes, but are there an infinite number of twin primes? This is one of the things that has yet to be figured out, and the Twin Prime Conjecture is asking this question.
Recently, a big jump was made in trying to prove this conjecture. So, I will post the article from News Scientist that describes the latest advancements on it.
http://www.newscientist.com/article/dn23644gameofproofsboostsprimepairresultbymillions.html#.UcxiPkIWHFI
This is often an odd concept to people, since math seems like it is all figured out. How could it be taught so definitively if there are still things to figure out? But, there are tons of conjectures out there that people are trying to prove.
One of them is called the Twin Prime Conjecture. You are probably aware that a prime number is a number that is only divisible by one and itself. There are an infinite number of prime numbers (click here to see a proof). Twin primes are two prime numbers that differ by two. For instance, 3 and 5 are twin primes because they are both prime and differ by two.
We know that there are an infinite number of primes, but are there an infinite number of twin primes? This is one of the things that has yet to be figured out, and the Twin Prime Conjecture is asking this question.
Recently, a big jump was made in trying to prove this conjecture. So, I will post the article from News Scientist that describes the latest advancements on it.
http://www.newscientist.com/article/dn23644gameofproofsboostsprimepairresultbymillions.html#.UcxiPkIWHFI
Saturday, July 6, 2013
The Law of Cosines
At some point during your high school math experience, whether it be during an Algebra II, Geometry, Trigonometry, or Precalculus class, you probably were presented with the Law of Sines and the Law of Cosines. You found that the sine and cosine ratio could only solve right triangles, and that you needed something more to solve nonright triangles. So, two formulas were slapped onto your textbook, and you had to apply them.
Each of the derived measurements are just rewritten forms of the sine and cosine ratios. For example, the cosine of C is that adjacent side over b, so multiplying both sides by b yields the measure of the adjacent side: b • cosC.
Let's look at the right triangle with hypotenuse c, and solve the Pythagorean Theorem. Since the Law of Cosines is similar to the Pythagorean Theorem, this might give us a start.
[a  (b • cosC)]^{2} + (b • sinC)^{2} = c^{2}
a^{2}  2abcosC + b^{2}cos^{2}C + b^{2}sin^{2}C = c^{2}
a^{2} + b^{2}cos^{2}C + b^{2}sin^{2}C  2abcosC = c^{2}
a^{2} + b^{2}(cos^{2}C + sin^{2}C)  2abcosC = c^{2}
^{} Wait a minute  what is inside the parentheses? We have cos^{2}C + sin^{2}C. But we proved earlier that that is always equal to one. So, we can substitute one in for that sum, and see where that takes us.
a^{2} + b^{2}(cos^{2}C + sin^{2}C)  2abcosC = c^{2}
a^{2} + b^{2}(1)  2abcosC = c^{2}
a^{2} + b^{2}  2abcosC = c^{2}
^{} And we end up with the Law of Cosines: a^{2} + b^{2}  2abcosC = c^{2}. I think that this proof is a pretty cool one, considering that it brings so many other neat identities into play.
Today, I am just going to focus on the Law of Cosines, and save the Law of Sines for a future post. This formula is one where you probably wondered why it worked. You might be asked to prove something using this formula, but how can you comfortably do that without being sure of the formula in the first place?
First, I need to lay a little foundation. Take a right triangle:
First off, we should know the Pythagorean Theorem. Click here for an explanation and proof of it.
Let's say we are dealing with the bottom left angle. The side opposite to this angle has a measure of 4, and will be referred to as the "opposite" side. The longest side has a measure of 5, and will be referred to as the "hypotenuse." The remaining side has a measure of 3, and will be referred to as the "adjacent" side, since it is adjacent to the angle.
Let's say we are dealing with the bottom left angle. The side opposite to this angle has a measure of 4, and will be referred to as the "opposite" side. The longest side has a measure of 5, and will be referred to as the "hypotenuse." The remaining side has a measure of 3, and will be referred to as the "adjacent" side, since it is adjacent to the angle.
Now, let's make sure we are on the same page with terminology. The sine of this angle is the opposite side over the hypotenuse. 4/5 is 0.8, so the sine of that angle is 0.8. The cosine of this angle is the adjacent side over the hypotenuse. 3/5 is 0.6, so the cosine of that angle is 0.6. Tangents will not be needed in this post, but it is the opposite over the adjacent. These ratios are commonly remembered by the SOH CAH TOA acronym.
There is a pretty cool identity found in the sine and cosine ratios. Take the two ratios that we just found for the 345 triangle:
sin(x) = 0.8
cos(x) = 0.6
Now, plug those into the following expression:
sin^{2}(x) + cos^{2}(x)
0.8^{2} + 0.6^{2}
0.64 + 0.36
1
Interestingly, this sum always turns out to be one. By using those ratios and the Pythagorean Theorem, you can prove that for all angles, the square of the sine plus the square of the cosine is one.
With this information, we can prove the Law of Cosines. The formula slightly resembles the Pythagorean Theorem, so we will try to keep that in mind when proving it.
0.8^{2} + 0.6^{2}
0.64 + 0.36
1
Interestingly, this sum always turns out to be one. By using those ratios and the Pythagorean Theorem, you can prove that for all angles, the square of the sine plus the square of the cosine is one.
With this information, we can prove the Law of Cosines. The formula slightly resembles the Pythagorean Theorem, so we will try to keep that in mind when proving it.
Each of the derived measurements are just rewritten forms of the sine and cosine ratios. For example, the cosine of C is that adjacent side over b, so multiplying both sides by b yields the measure of the adjacent side: b • cosC.
Let's look at the right triangle with hypotenuse c, and solve the Pythagorean Theorem. Since the Law of Cosines is similar to the Pythagorean Theorem, this might give us a start.
[a  (b • cosC)]^{2} + (b • sinC)^{2} = c^{2}
a^{2}  2abcosC + b^{2}cos^{2}C + b^{2}sin^{2}C = c^{2}
a^{2} + b^{2}cos^{2}C + b^{2}sin^{2}C  2abcosC = c^{2}
a^{2} + b^{2}(cos^{2}C + sin^{2}C)  2abcosC = c^{2}
^{} Wait a minute  what is inside the parentheses? We have cos^{2}C + sin^{2}C. But we proved earlier that that is always equal to one. So, we can substitute one in for that sum, and see where that takes us.
a^{2} + b^{2}(cos^{2}C + sin^{2}C)  2abcosC = c^{2}
a^{2} + b^{2}(1)  2abcosC = c^{2}
a^{2} + b^{2}  2abcosC = c^{2}
^{} And we end up with the Law of Cosines: a^{2} + b^{2}  2abcosC = c^{2}. I think that this proof is a pretty cool one, considering that it brings so many other neat identities into play.
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