Question 8

8. In class we showed that √2 is an irrational number; that is, there is no fraction ab, where a and b are whole numbers and a/b=√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 a and b are co-prime (they share no factors). In particular, this means that a and b CANNOT both be even. This gives us exactly 3 possibilities for the parity of a and b. If we rewrite a/b=√2 as a=b√2, 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: 

1) 2 is 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. 

2) 5 is the solution to x-5=0. So it too is an algebraic number. Duh.

3) 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. 

  

So, let's first start with 8a, and let's show that 7 + √2 is an irrational number.

First, Ben reminded me to assume that 7 + √2 is equal to a/b.

So:

7 + √2 = a / b

7b + √2b = a

√2b = a - 7b
Now, let's think about it. Is there anyway for an irrational number multiplied to an integer to create an integer? No! So this means that any result on the left side of the equation has to be an irrational number. Now, let's look at the right side. What does 7 times any rational number make? Another rational number. So, is there anyway for a rational number minus another rational number to create an irrational number? No! Because any rational subtracted from another rational number will create a rational number.
This means that the left side is an irrational number and the right side is an rational number. There is no way for number to be both rational and irrational, so 7 +√2 is an irrational number.

Then for 8b. How do we know whether or not 7 +√2 is an algebraic number? Well, we'd first have to define what an algebraic number is. According to the backgound, algebraic numbers are numbers who are solutions to polynomial equations with rational numbers for their coefficients. So, there are two requirements, algebraic numbers must be a solution to an algebraic equation, and this equation must only have rational numbers like fractions and whole numbers.
So, to show that this is possible, let's work backwards and assume there is a problem in which the solution is:
x = 7 +√2
x - 7 = √2
(x-7)²= 2
(x-7)² - 2 = 0
x² - 14x + 49 -2 = 0
x² - 14x + 47 = 0
I was able to work backwards and produce an algebraic equation that has only rational numbers and, because, I worked backwards, to make sure that 7 + √2 is the solution to the equation then 7 + √2 has to be an algebraic number because I've found an algebraic equation that has only rational numbers and in which 7 + √2 is the solution.

Bonus: Prove that √3 is irrational
To prove that √3 is irrational, I first have to assume that it is rational.
If that is the case then there will be a fraction such that two non-common factors a and b divided by each other that is equal to the √3:
√3 = a/b
(√3 = a/b)²
3 = a² / b²  
3b² = a²
So, let's consider all the possibilities,
If a is odd and b is even then:
3(even #)² = (odd #)²
3 x even # = odd #
even # = odd #
This is not possible, a number can't both be odd and even at the same time, so this is not possible.
What if a is even and b is odd, then:
3(odd #)² = (even #)²
3 x odd # = even #
odd # = even #
The same problem happens again and, like before, a number can't both be odd and even at the same time, so this is not possible.
Alright, so what if both a and b were even, this there will be no problem then right?
3 (even #) = (even #)
3 x even # = even #
even # = even #
However, if two numbers are even one can always divide them by 2 and so our initial assumption that a/b have no common factor is wrong, because if they do, they can further be simplified. I can then imply that a and b cannot be both even.
Now, what if both a and b were odd?
Well, if they both were odd then I would be able to rewrite them like this:
a = 2s + 1
b = 2t + 1
The s and t are required to be integers, because this is the only way to ensure that a and b are integers.
Now, let's plug these in for a and b:
3(2t+1)² = (2s+1)²
3(4t² + 4t + 1) = (4s² + 4s + 1) 
12t² + 12t + 3 = 4s² + 4s + 1
(12t² + 12t + 2 = 4s² + 4s) x 1/2
6t² + 6t + 1 = 2s² + 2s
If I look into this, any integer that I plug in for t on the left will create an odd number, and because there's not a +1 on the right side of the equation, any integer I plug in for s will create an even number. Like before, there's no way for a number to be both odd and even at the same time so a and b can't both be odd.
So, there are no combinations of a and b (made of non-common factors) such that a/b is equal to √3. If there are no such cases, then that means the √3 is irrational.

Question 7

7. Use Geometric Algebra to show that a2-b2=(a+b)(a-b). You can use the distributive rule in your demonstration. Do NOT use negative numbers.

First, let's prove it algebraically:

a² – b² = (a-b)(a+b)

a² – b² = a² – ab + ab – b²

a² – b² = a² – b²

Now, that I've seen that, let's try to prove it using geometric algebra. Let's first start with a square, with sides a.

Then, on one side, let's add b, and on the other let's go backwards b.

Once, this has been done, let's look at the part where I actually want to find the area (the orange part):

Why do I only look at this part? Because the bottom purple part has been taken out, we want to look at the side (a-b) only and not a, as a whole. Therefore, the bottom purple rectangle is no longer important. If this is the case, why is the rectangle on the right important? Because we want to look at a + b , and the rectangle on the right along with a is representative of this.

Now, let's remove the bottom part, and concentrate on what we really want to look at:

The top and bottom side is now: a + b, and the left and right side is now representative of a - b. But how is their area a2 - b2?

To make this easier, Gabe showed me how to move the rectangle with top side b around, and put it on the bottom.
The end result of the move should look like this:

Now that I've moved it, we can see that there is now a whole side of a on the left and on the top. But, wait, how did this work out? Well, when I moved it, I took the extra part of b from a + b away, so it's like a + b - b, and now we have only a on the top. Then, by moving this part of b to the part of a that I took it out of, I replaced the missing b on one side - in a way, I did a - b + b. 

Now I have a whole a on both the top side and the left side, I know that if the bottom and right side were also filled, it would be like the original square that I had way at the beginning, which has an area of::

a x a = a²

But, this square is not "whole" it has a hole that's in the shape of square in the middle with sides b. How do I know that the sides are b? Well, from the move and from knowledge of before. Before the move, the now-bottom side has a length of a - b. So after the move, the bottom side should still have a length of a - b. If  the top is length a, and the bottom is length a - b, then the difference between them is a length of b, which I have labeled. I know the left side of the small square is also length b because before I moved it, the length was "b", so after the move, the length should remain the same. Therefore, I know that the hole on the corner of square a is a small square made from two sides of b.

So, to find the area of the orange part only, I have to take the area of the large square a, and take out the area of a small square b. I already know the area of the large square a to be a², so I only need to find the area of the small square b.

The area of the small square b is:

b x b = b²

Therefore, the area of the orange rectangles which is representative of (a-b)(a+b) combined is:

a² – b² 

So, if the area of the orange rectangles is representative of  (a-b)(a+b) then:

(a-b)(a+b) = a² – b²

Question 6

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.

To solve this problem, I have to first draw a circle, and then a triangle on half of it. The picture should look like the picture below:

I then named each corner of the triangle to be A,B, and C. Then, I put a center D on the circle and connected in to point B. Because, D is at the center of the circle - it means that any point on the circle from D to that point is the radius and they are all equal. This means:

AD = CD = BD

Then, let's look at the two smaller triangles created by BD.

Triangle ABD is an isoceles triangle - it has two sides that are equal. How do I know this? Two of it sides are AD, and BD, which I have stated earlier to be the same because they are both the radius of the circle. That means that the angles opposite of these two lines are also the same meaning that the orange angle (let's call it x) and red angle (let's call it y) is equal. x = y

I know that that the sum of all the angles in a triangle is = 180, so:

x + y + z = 180

z = 180 - x - y

Then, I repeated the same steps for the other triangle. Triangle CBD is also a isoceles triangle. I know this because has sides BD and CD which I have also stated earlier are equal to each other. This then means that the angles opposite of BD, and CD - purple angle l and brown angle n are equal to each other. l = n.

Then, I know that the sum of all the angles in a triangle is = 180, so:

l + m + n = 180

m = 180 - l - n

Looking at the picture, I know that triangle ABC is produced from "adding" these two smaller triangles together, so I can also add these two equations together:

(z = 180 - x - y ) + (m = 180 - l - n)

z + m = 360 - x - y - l - n

Then, what do I do next? I have to 6 variables and only three equations. Well, I know that l = n and x = y so let's get rid of two variables by replacing them with their equivalent. So, for my purposes, I will get rid of n and replace it with l, and get rid of x, and replace it with y:

z + m = 360 - y - y - l - l

z + m = 360 -2y - 2l

Then, I know angles z + m  180. Why? Because the two of them combined makes a straight line, and I know that straight lines are 180 degrees. So, let's replace z + m with 180 in the equation:

180 = 360 - 2y - 2l

2y + 2l + 180 = 360

2y + 2l = 180

y + l = 90

Angle y + angle l = angle b in the larger triangle, triangle ABC (I obtain this from looking at the picture). 

If angle y + angle l = 90, and angle y + angle l = angle b then:

angle b = 90.

If angle b is 90 degrees then an angle inscribed in a semi circle is a right angle.

Question 5

5. Use Egyptian Multiplication to compute 571 x 73. Then express your answer in binary notation.

To do Egyptian Multiplication, I have to make a table (like the one above). On the left side, I start with 1, and double each number until the number is just below 571 and it can no longer be doubled because if it is, it will be larger than 571 (in this case, 1024 > 571). On the right side, I start with 73 and double each number until the table is completely filled out and each number on the left has a corresponding number on the right.

After that, I have to determine which numbers on the left side of the table add up to 571 (the boxed numbers).
I choose:
512 + 32 + 16 + 8 + 2 + 1 = 571
Then, to determine what 571 x 53 is equal to, I add the numbers on the right side of the table that corresponds to the ones on the left that I chose: (For example, the number corresponding to 512 is 37376, and the number that corresponds to 32 is 2336, etc.)
37376 + 2336 + 1168 + 584 + 146 + 73 = 41,683
This number of 41,683 = 571 x 73.
To turn 41,683 into a binary number, I have to:
41683 / 2 = 20841 with a remainder of 1
20841 / 2 = 10420 with a remainder of 1
10420 / 2 = 5210 with a remainder of 0
5210 / 2 = 2605 with a remainder of 0
2605 / 2 = 1302 with a remainder of 1
1302 / 2 = 651 with a remainder of 0
651 / 2 = 325 with a remainder of 1
325 / 2 = 162 with a remainder of 1
162 / 2 = 81 with a remainder of 0
81 / 2 = 40 with a remainder of 1
40 / 2 = 20 with a remainder of 0
20 / 2 = 10 with a remainder of 0
10 /2 = 5 with a remainder of 0
5 / 2 = 2 with a remainder of 1
2 / 2 = 1 with a remainder of 0
1 /2 = 0 with a remainder of 1
Then, I read the sequence of remainders upwards to the top to determine the binary number. So, the binary number of 41,683 is: 1010001011010011.

Question 4

4. Generate a cubic:

ax3+bx2+cx+d

where a and b are nonzero integers so that your cubic is the un-depressed version of the cubic in Chapter 6: x3+6x-20. Next, use the solution to the depressed cubic in Chapter 6 to find a solution to your cubic.

To do this, I have to start with the general form:

let x = y - b/3a

ay³ + by² + cy + d = 0
Ben then reminds me to obtain another way this equation is written from Journey through Genius
ay³ + 0y² + (c -  (b²/3a)y + (d+(2b³/27a²)-(bc/3a)) = 0
So then I compare this to the depressed cubic x³+6x-20=0 that I'm un-depressing
ay³ = x³
(c -  (b²/3a)y = 6x
(d+(2b³/27a²)-(bc/3a)) = -20
If, for just a moment, I ignore the difference in variables and assume that x and y are equal because I do have an equation to convert between them (x = y - b/3a) then:
ax³ = x³; so a = 1

(c -  (b²/3a)x = 6x; (c -  (b²/3a) = 6 - another way to write this is: c = 6 + b²/3a
(d+(2b³/27a²)-(bc/3a)) = -20; plug that value of "c" and "a" that we've figured out into this equation:
d+(2b³/27a²)-(b(6 + b²/3a)/3a)) = -20
d+(2b³/27(1)²)-(b(6 + b²/3(1))/3(1))) = -20
( d+(2b³/27)-(b(6 + b²)/3) = -20) 27 - multiply both sides by 27 to get rid of the fractions:
27d + 2b³ - 9b(6 + b²) = -540
27d + 2b³ - 9b³ - 54b = -540
27d - 7b³ - 54b = -540 - at this point, I don't have another equation to solve for d and b, but I can approximate their value by trial and error. I also know that b has to be a non-zero integer. But, other than this limitation, b can be any number as long as it satisfies the equation. I then decided to try b = 3 because 3 is a non-zero integer and I can easily find the value of 3³.
So:
27d - (7)3³ - 54(3) = -540
27d - 189 -  162 = -540
 27d - 351 = -540
27d = -189
d = -189/27 = -7
Then, I still have to solve for the actual value of c:
c = 6 + b²/3a
c = 6 + 3²/3(1)
c = 6 +9/3
c = 6 + 3
c = 9
Now, I that know all the values of a, b, c, and d, I can plug this back into the general form:
ay³ + by² + cy + d = 0, which can be rewritten as:
ax³ + bx² + cx + d = 0
x³ + 3x² + 9x - 7 = 0

To find the solution to my cubic I need to use this equation:
x = y - b/3a, which I can rewrite as:
x₁ = x₂ – b/3a; x₁ is the solution to my cubic and x₂ is the solution to x³+6x-20

Dunham had found in Journey through Genius that the solution to x³+6x-20 = 0 is that x = 2

Therefore, x₂ = 2, and I already know b and a from doing the problem earlier, b = 3 and a=1.

So:

 x₁ = 2 – (3/3(1))

 x₁ = 2 - (3/3)

x₁ = 2 - 1

x₁ = 1

Question 3

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.

Other than using calculus, the value of π can be approximated using polygons. Traditionally, Archimedes, who established this method of approximation used polygons inscribed and circumscribed around a circle to establish an lower and upper bounds on the value of pi. But, in this post I will only approximate pi using inscribed polygons. So, how does this approximation work? First, let's find the approximation of π using hexagon and then a 12-gon before we get to a 24-gon.

First, make a circle:

 Then I draw a hexagon inside the circle.

 

After the hexagon is drawn, I can draw a line across the circle that cuts through the hexagon and connects two opposite corners. This line that cuts through the circle is the diameter.

I can then finish "cutting" the hexagon by connecting the last two pair of opposites like this picture. From the previous picture, we can see that all of these "cuts" is the circle's diameter. It can be anything, but in this problem, I will assume that the diameter is 2 and because  radius = diameter / 2, r = 2/2 = 1. Because the triangles created are equilateral triangles we know that all the sides of the triangles and the hexagon is equal to the radius which is equal to 1.
We can then approximate pi using this:

C = 2πr

π = C / 2r

But, we can't find the area without using pi, so we will approximate it using the perimeter of the hexagon.

The perimeter of the hexagon is: the number of the sides x the length of the sides, in this case it's:

Perimeter = 6 x 1 = 6.

So,

π is approximately = 6 / 2(1) = 6/2 = 3

Now, to improve this approximation let's add more lines and cut each of the triangles in half to prepare and make a 12-gon(like the picture below).

Then, connect the line in the middle to the corners to make this picture below. I've made the 12-gon.

 
Here, in this picture, I take 1 slice of the equilateral triangle from before out so that it is easier to see the length of the sides to this 12-gon.This picture shows the variables of all the things we are missing. N is a length of one side of the 12-gon, this is what I ultimately want to find because with this length I can find the perimeter of the hexagon and use it. However, first, I'd have to find the length of m, and the only way to do that is find the length of L.

Because L + m = 1 (it is the radius), once I find L then 1 - L = m.

 
So, to make this problem even easier, I've split the two triangles and made them larger so it is easier to see. This triangle is the one in which we will use to determine L. We know that r is the radius and so it is 1. We then know that h = 1/2r (because the equilateral triangle from the hexagon was cut into 2) so, in this case, h = ½.

Then we use the pythagorean theorem:

½² + L² = 1²

¼ + L² = 1

L² = 1 – ¼

L² = ¾

L = √3/2

Then, to find n:

First, we find m: L + m = 1, so m = 1 -  √3/2 = (2-√3)/2

I know h = ½

So, let's use the pythagoren theorem:

½² + (2-√3/2)² = n²

¼ + (4-4√3+3)/4 = n²

                                           (8-4√3)/4 = n²

                                           2-√3 = n²

                                           √(2-√3) = n

Now that I have n, which is the length of one side of the 12-gon, I can find the perimeter of the 12-gon to approximate pi:

Perimeter = 12 x √(2-√3) = 12√(2-√3)

I can then approximate pi using this:

C = 2πr
Assuming C is approximately the perimeter of the 12-gon:

π = Perimeter / 2r = 12√(2-√3)/2 = 6√(2-√3) = 3.10582


Then, to approximate pi from a 24-gon, cut the triangles of the 12-gon in half and repeat the procedure again. The result will look like this:

Then I split the triangles like before, so that I can determine the length of one side:
After a similar math process to the 12-gon, I determine that one side of the 24-gon is:
√(2-(√2+(√3)))
So, to approximate pi, I then need to find the perimeter of the 24-gon
Perimeter = 24 x √(2-(√2+(√3))) = 24√(2-(√2+(√3)))
I can then approximate pi using this:

C = 2πr
Assuming C is approximately the perimeter of the 24-gon:

π = Perimeter / 2r = 24√√(2-(√2+(√3)))/2 = 12√(2-(√2+(√3))) = 3.1326286

Bonus - Go all the way to 96. This makes you current with Archimedes.

The 24-gon is then cut again to create a 48-gon, and I can use the previous knowledge of the previous length to figure out that the length of one side of it is:

 √(2-(√2+(√2+(√3))))

Then, knowing this, I can "cut"the triangles in the 48-gon in half again to create a 96-gon. And, using the information from the previous polygon, I determined the length of one side of the 96-gon to be:

 √(2-(√2+(√2+(√2+(√3)))))

Then, to approximate pi, I have to first find the perimeter of this polygon:

Perimeter = 96 x  √(2-(√2+(√2+(√2+(√3))))) = 96 √(2-(√2+(√2+(√2+(√3)))))

I can then approximate pi using this:

C = 2πr
Assuming C is approximately the perimeter of the 24-gon:

π = Perimeter / 2r = 96 √(2-(√2+(√2+(√2+(√3)))))/2 = 48√(2-(√2+(√2+(√2+(√3))))) = 3.14103195

Question 2

2. Find all factors of 496 and prove that 496 is a perfect number.

All the factors of 496 are:
1, 2, 4, 8, 16, 31, 62, 124, 248, and 496.
How do I know that these are ALL of the factors of 496? I used a factoring tree!

Using the factoring tree, I can determine all of the possible factors of 496.
I know:
2 x 248 = 496
2 x 2 x 124 = 496; 4 x 124 = 496
2 x 2 x 2 x 62 = 496; 8 x 62 = 496
2 x 2 x 2 x 2 x 31 = 496; 16 x 31 = 496
How do I know that two is a factor of 496 and how do I know to stop at 31? I know that as long as a number is even, it can be factored over and over using the number two. I also know that 31 is a prime number using the last problem and because it is no longer factorable, these are all the possible factors of 496.

496 is a perfect number. How do I know this? First, I have to understand what a perfect number is. A perfect number is whole, positive integer that is equal to the sum of all of its factors, excluding itself.
                For example, 6 is a perfect number.
                I know that the factors of 6 are: 1, 2, 3, and 6.
                1 + 2 + 3 = 6
Then, what about 496?
1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496
(1 + 2 + 4 + 8 +16  = 31)
31 + 31 + 62 + 124 + 248 = 496
(31 + 31 = 62)
62 + 62 + 124 + 248 =  496
(62 + 62 = 124)
124 + 124 + 248 = 496
(124 + 124 = 248)
248 + 248 = 496

Bonus - Do the same for 8.128 and 8,589,869,056.
Using the same method, I found all of the factors of 8,128 are:
1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064 and 8128
So, is it a perfect number? Let's add all of the factors except for itself together:
1+2+4+8+16+32+64+127+254+508+1016+2032+4064 = 8128.
This works! Therefore, 8128 is a perfect number.
What about 8,589,869,056? What are all of its factors? The factors of 8,589,869,056 are:
1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131071, 262142, 524284, 1048568, 2097136, 4194272, 8388544, 16777088, 33554176, 67108352, 134216704, 168433408, 536866816, 1073733632, 2147467264, 4294934528, 8589869056
So, is it a perfect number?
Let's add!
1+2+4+8+16+32+64+128+256+512+1024+2048+4096+8192+16384+32768+65536+131071+262142
+524284+1048568+2097136+4194272+8388544+16777088+33554176+67108352+134216704+ 168433408+536866816+1073733632+2147467264+4294934528 = 8,589,869,056
This is true so it means that 8,589,869,056 is a perfect number!

Question 1

1. Consider the sequence 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+1 = 3

2 x 3+1 = 7

2 x 3 x 5+1 = 31

2 x 3 x 5 x 7+1 = 211

2 x 3 x 5 x 7 x 11+1 = 2,311
and etc.

The first member of this sequence that is not a prime is:

2 x 3 x 5 x 7 x 11 x 13+1 = 30,031; it can be made by multiplying 59 x 509

So.. how do I know this that this is the first member of the sequence that is not a prime? First, I would need to define what a prime number is. A prime number is any number that has no other positive, whole number as divisors except 1 and itself. 

Using this information, I know 3, 7, and 31 are prime numbers. But, how did I figure that 2,311 is a prime number and 30,031 isn't?

First, using information from Dr. Math's forum: http://mathforum.org/library/drmath/view/54371.html, I found out that of all the divisors (factors) of a number, one of the two divisors has to be less than the square root of the number.

                  For example, 12 is the number that I want to find factors for. 

                                              The √12 is approximately 3.46410

                                              Factors of 12 includes: 

                                              1 x 12 = 12, here 1 < 3.46410

                                              2 x 6 = 12, here 2 < 3.46410

                                              3 x 4 = 12, here 3 < 3.46410

So, from this, I know that any pair of divisors of 2311 has to have one that is less than the √2311 = 48.072 and any pair of divisors of 30,031 has to have one that is less than √30,031 = 173.2945.

Then, I can test them using divisibility. If they are prime, any number less than those would not give a remainder of 0. I figured that because all of the numbers made by this sequence (due to the +1 at the end) are not even numbers, I don't have to test 2 or any number like 4, 6, 8, 10, etc. because only even numbers would be divisible by these numbers to give a remainder of 0. Not only that, from knowledge in elementary school, I can go on further to say I don't have to test non-prime numbers because anything that's divisible by 9 can be divided by 3. So, I only have to divide these numbers by prime numbers that are smaller than their square roots.

                  For example, 2311 - √2311 = 48.072 so the divisors (if there are any) has to be less than 48 and prime.

                    Then, using my knowledge of prime numbers, I can figure out and list the ones that are < 48, they are:

3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

                                           2311/3 = 770.3333...

                                           2311/5 = 462.2

                                           2311/7 = 330.1428...

                                           2311/11 = 210.09...

                                           2311/13 = 177.769...

                                           2311/17 = 135.941...

                                           2311/19 = 121.632...

                                           2311/23 = 100.478...

                                           2311/29 = 79.689...

                                           2311/31 = 74.548...

                                           2311/37 = 62.459...

                                           2311/41 = 56.365...

                                           2311/43 = 53.744...

                                           2311/47 = 49.170...

From this list, I know that 2311 is a prime number because it has no prime divisors < 48.072 that gives a remainder of 0.

I then repeat this process for 30,031 and found that it's not prime because it can be divided using 59 and the answer will be 509 with 0 as a remainder. This means that 59 and 509 (in fact, 59 x 509 = 30,031) are factors of 30,031 and it is NOT a prime number.