Z4 = 24-1 = 23 = 8 =
Although it appears that Zn = 2n-1(Z), we must check other values to determine that the equation does stay true.
When n = 15
Z15 = (Z4)(Z4)(Z4)(Z3) = =
Verifying with Zn = 2n-1(Z),
Z15 = 215-1 = 214 = 16384 =
When n = 9
Z9 = (Z4)(Z4)(Z) = =
Verifying with Zn = 2n-1(Z),
Z9 = 29-1 = 28 = 256 =
Testing when n = 13
Z13 = (Z4)(Z4)(Z4)(Z) = =
With this evidence, we can come to the conclusion that Zn = 2n-1(Z). However we must take into consideration that Z = (X + Y). This ultimately leads to:
(X + Y)n = 2n-1(X + Y)
Next, we’ll consider A. A is a value and can be represented by A = aX. The value of (a) is always a constant, and we’ll use the value of X from the previous work. Using this information, we’ll find the values of A2, A3, A4, and An.
To find a pattern within the values, we’ll test using various values of (a).
Remember that X = and A = aX.
When a = 2,
A2 = 2 2 = =
A3 = =
A4 = =
When a = 4
A2 = 4 4 = =
A3 = =
A4 = =
When a = 6
A2 = 6 6 = =
A3 = =
A4 = =
When a = 8
A2 = 8 8 = =
A3 = =
A4 = =
When a = 10
A2 = 10 10 = =
A3 = =
A4 = =
When a = 5
A2 = 5 5 = =
A3 = =
A4 = =
When a = 3
A2 = 3 3 = =
A3 = =
A4 = =
Because A = aX, we must remember that X = 2n-1(X). In finding An, we must find how the value of (a) changes. We can see this by looking at the results of An and how it corresponds to (a).
Since An means (aX)n, we can see that as (a)n (X)n. Since Xn = Xn = 2n-1(X), we can consequently combine the two equations which means that An = 2n-1(X) (a)n
We can verify this by testing the previous values with the equation.
When a = 5
A2 = A3 = A4 =
Now using the equation An = 2n-1(X) (a)n
A4 = 54 =
When a = 10
A2 = A3 = A4 =
Now using the equation An = 2n-1(X) (a)n
A4 = 104 =
When a = 8
A2 = A3 = A4 =
Now using the equation An = 2n-1(X) (a)n
A4 = 84 =
From this evidence, we can conclude that An equals 2n-1(X) (a)n.
Next, we’ll consider B. B is a value and can be represented by B = aY. The value of (b) is always a constant, and we’ll use the value of X from the previous work. Using this information, we’ll find the values of B2, B3, B4, and Bn.
To find a pattern within the values, we’ll test using various values of (b).
Remember that Y = and B = bY.
When b = 2,
B2 = 2 2 = =
B3 = =
B4 = =
When b = 4
B2 = 4 4 = =
B3 = =
B4 = =
When b = 6
B2 = 6 6 = =
B3 = =
B4 = =
When b = 8
B2 = 8 8 = =
B3 = =
B4 = =
When b = 10
B2 = 10 10 = =
B3 = =
B4 = =
When b = 5
B2 = 5 5 = =
B3 = =
B4 = =
When b = 3
B2 = 3 3 = =
B3 = =
B4 = =
Because B = bY, we must remember that Y = 2n-1(Y). In finding Bn, we must find how the value of (b) changes. We can see this by looking at the results of Bn and how it corresponds to (b).
Since Bn means (bY)n, we can see that as (b)n (Y)n. Since Yn = 2n-1(Y), we can consequently combine the two equations which means that Bn = 2n-1(Y) (b)n
We can verify this by testing the previous values with the equation.
When b = 5
B2 = B3 = B4 =
Now using the equation Bn = 2n-1(Y) (b)n
B4 = 54 =
When b = 10
B2 = B3 = B4 =
Now using the equation Bn = 2n-1(Y) (b)n
B4 = 104 =
When b = 8
B2 = B3 = B4 =
Now using the equation Bn = 2n-1(Y) (b)n
B4 = 84 =
From this evidence, we can conclude that Bn equals 2n-1(Y) (b)n.
Next we’ll take into consideration the equation (A+B)n.
Looking at (A+B), we can see that it translates to (aX+bY) meaning that (a) and (b) are still constants. Since (a) and (b) are not necessarily always the same values, we must also take into concern and test for different values.
When a = 2 and b = 2
(A+B)2 = = = =
(A+B)3 = =
(A+B)4 = =
When a = 4 and b = 4
(A+B)2 = = = =
(A+B)3 = =
(A+B)4 = =
When a = 6 and b = 6
(A+B)2 = = = =
(A+B)3 = =
(A+B)4 = =
When a = 8 and b = 8
(A+B)2 = = = =
(A+B)3 = =
(A+B)4 = =
Now we’ll take into consideration values for when (a) and (b) are different.
When a = 2 and b = 4
(A+B) 2 = = = =
(A+B) 3 = =
(A+B)4 = =
When a = 8 and b = 6
(A+B)2 = = = =
(A+B)3 = =
(A+B)4 = =
To find (A+B)n, we’ll look at the data above. The information above suggests that An+Bn = (A+B)n. However, we must test this with data from above to verify and confirm if it’s true. While testing, we must remember to test for values of a and b that are both the same and different.
When a = 2 and b = 2
(A+B)2 = (A+B)3 = (A+B)4 =
Verifying using (A+B)n = An + Bn
(A+B)4 = A4 + B4 = + =
When a = 4 and b = 4
(A+B)2 = (A+B)3 = (A+B)4 =
Verifying using (A+B)n = An + Bn
(A+B)4 = A4 + B4 = + =
When a = 6 and b = 6
(A+B)2 = (A+B)3 = (A+B)4 =
Verifying using (A+B)n = An + Bn
(A+B)4 =A4 + B4 = + =
When a = 8 and b = 8
(A+B)2 = (A+B)3 = (A+B)4 =
Verifying using (A+B)n = An + Bn
(A+B)4 = A4 + B4 = + =
When a = 2 and b = 4
(A+B)2 = (A+B)3 = (A+B)4 =
Verifying using (A+B)n = An + Bn
(A+B)4 = A4 + B4 = + =
When a = 8 and b = 6
(A+B)2 = (A+B)3 = (A+B)4 =
Verifying using (A+B)n = An + Bn
(A+B)4 = A4 +B4 = + =
Ultimately, from the information above and the following verification, we can conclude that the formula for (A+B)n is (A+B)n = An + Bn.
Finally, we’ll consider M. The value for M is. Although it is given that M = A + B, we must come to this conclusion using data and calculated information. Hence, since A = aX and B = bY, we must remember that the values for a and b are constants and we must consider that can be both the same and different. For this reason, we’ll test multiple values for a and b to come to the conclusion that M =.
When a = 2 and b = 2
Using the formula for M, this would translate to which equals.
Using the equation M = A + B
M = = =
When a = 4 and b = 4
Using the formula for M, this would translate to which equals.
Using the equation M = A + B
M = 4 + 4 = + =
When a = 2 and b = 4
Using the formula for M, this would translate to which equals
Using the equation M = A + B
M = = =
When a = 8 and b = 6
Using the formula for M, this would translate to which equals
Using the equation M = A + B
M = = + =
From this, we can conclude that the equation M = A + B is true because of the values tested and confirmed above.
This leads us to consider M2. Since M = , then M2 = .
To find the simplified value of M2, multiply it out. This translates to .
When multiplying matrices, one must add up the products of corresponding rows and columns. However, dimensions must match. If two matrices were represented by and , then the multiplied matrix would equal .
When using this information regarding M2, we see that
M2 =.
Multiplying this out would result in:
After multiplying this out, the result is the 2x2 matrix
Now that we have the simplified version of M2, we can use it to show that M2 = A2 + B2. However, since A = aX and B = bY, we must remember that the values for a and b are constants and we must consider that can be both the same and different. For this reason, we’ll test multiple values for a and b to come to the conclusion that
M2 =
When a = 2 and b =2
Using the formula for M, this would translate to
=
Verifying using M2 = A2 + B2
M2 = + =
When a = 8 and b = 8
Using the formula for M, this would translate to
=
Verifying using M2 = A2 + B2
M2 = + =
When a = 2 and b = 4
Using the formula for M, this would translate to
=
Verifying using M2 = A2 + B2
M2 = + =
When a = 8 and b = 6
Using the formula for M, this would translate to
=
Verifying using M2 = A2 + B2
M2 = + =
Thus, we can conclude that M2 = A2 + B2 because all of the values above have been the same whether it was the variable version of M2, , or plugging into the equation M2 = A2 + B2
Finally, we’ll look at the value of Mn. To find the equation for Mn, we can look at the equation of M2. When comparing Mn to M2 we can see that n=2. Using the connection established, the equation M2 = A2 + B2 can be translated to Mn = An + Bn. However, since M is a matrix expressed in terms of values of (a) and (b), matrix Mn must also be expressed in terms of (a) and (b). Since there is a connection between M2 and Mn, being that n=2, we can connect the two equations once again. Since the previous equation for (A+B)n came to be An + Bn, this reinforces that Mn = An + Bn. However, relating this back to the original matrix form of M, we must consider that the original formula for M was M =. Therefore, Mn must equal. However, this translates to equal the same as An + Bn because it is the same as Mn = (an)(2n-1)(X) + (bn)(2n-1)(Y). Nonetheless we must still test this with values to verify and confirm, and once again, we must take into consideration values of (a) and (b) that are both the same and different, and we can use both equations to make sure.
When a = 2 and b = 2
Using the equation Mn = (an)(2n-1)(X) + (bn)(2n-1)(Y).
M2 = (22)(2) + (22)(2) =
Verify using the Mn =
M2 = = =
When a = 8 and b = 8
Using the equation Mn = (an)(2n-1)(X) + (bn)(2n-1)(Y).
M2 = (82)(128) + (82)(128) =
Verify using the Mn =
M2 = =
When a = 8 and b = 6
Using the equation Mn = (an)(2n-1)(X) + (bn)(2n-1)(Y).
M2 = (82)(128) + (62)(32) =
Verify using the Mn =
M2 = =
From this, we can consequently conclude that Mn = because when plugged in for various values of (a) and various values of (b). The equation confirms and verifies to be true because we verified it using two equations that both equal
Mn = An + Bn.
Ultimately, the nine goals were reached based off calculations of the initial values of X and Y and all values and calculations were proven using various equations to verify and confirm