CALCU-5. For any positive integer n, an idempotent

5. For any positive integer

n

, an idempotent in

\mathbb{Z}_{n}

is an element

e

such that

e^{2}=e

. For example 0 and 1 are always idempotents in

\mathbb{Z}_{n}

if

n \geq 2

, and 3 and 4 are idempotents in

\mathbb{Z}_{6}

. (a) Find the idempotents in each of

\mathbb{Z}_{12}, \mathbb{Z}_{24}, \mathbb{Z}_{36}

and

\mathbb{Z}_{60}

. (If you simply list them correctly, you receive full marks without needing to provide a justification.) (b) Use part (a) or otherwise to evaluate

2^{100}

and

3^{100}

in each of

\mathbb{Z}_{12}, \mathbb{Z}_{24}

,

\mathbb{Z}_{36}

and

\mathbb{Z}_{60}

. Show your working. (c) Suppose that it is 10:24 am on a Monday. What time and day will it be after

2^{100}

minutes have elapsed? Show your working.

Solution by Steps

step 1

To find idempotents in

\mathbb{Z}_{n}

, we need to find all elements

e

such that

e^2 \equiv e \mod n

step 2

For

\mathbb{Z}_{12}

, we test each element from 0 to 11 to see if it satisfies the idempotent property

step 3

The idempotents in

\mathbb{Z}_{12}

are found to be 0, 1, 4, and 9

step 4

For

\mathbb{Z}_{24}

, we test each element from 0 to 23 to see if it satisfies the idempotent property

step 5

The idempotents in

\mathbb{Z}_{24}

are found to be 0, 1, 4, 9, 16, and 21

step 6

For

\mathbb{Z}_{36}

, we test each element from 0 to 35 to see if it satisfies the idempotent property

step 7

The idempotents in

\mathbb{Z}_{36}

are found to be 0, 1, 4, 9, 16, 25, and 36

step 8

For

\mathbb{Z}_{60}

, we test each element from 0 to 59 to see if it satisfies the idempotent property

step 9

The idempotents in

\mathbb{Z}_{60}

are found to be 0, 1, 4, 9, 16, 25, 36, and 49

Answer

Idempotents in

\mathbb{Z}_{12}

are 0, 1, 4, 9; in

\mathbb{Z}_{24}

are 0, 1, 4, 9, 16, 21; in

\mathbb{Z}_{36}

are 0, 1, 4, 9, 16, 25, 36; in

\mathbb{Z}_{60}

are 0, 1, 4, 9, 16, 25, 36, 49.

Key Concept

Finding idempotents in

\mathbb{Z}_{n}

Explanation

An idempotent in

\mathbb{Z}_{n}

is an element

e

such that

e^2 \equiv e \mod n

. We find these by testing each element within the set.

---

step 1

To evaluate

2^{100}

and

3^{100}

in

\mathbb{Z}_{n}

, we can use the idempotents found in part (a)

step 2

Since

2^2 = 4

and

3^2 = 9

are idempotents,

2^{100} = (2^2)^{50} = 4^{50} \equiv 4 \mod n

and

3^{100} = (3^2)^{50} = 9^{50} \equiv 9 \mod n

, where

n

is the modulus

step 3

We apply this to each modulus: for

\mathbb{Z}_{12}

,

2^{100} \equiv 4

and

3^{100} \equiv 9

step 4

For

\mathbb{Z}_{24}

,

2^{100} \equiv 4

and

3^{100} \equiv 9

step 5

For

\mathbb{Z}_{36}

,

2^{100} \equiv 4

and

3^{100} \equiv 9

step 6

For

\mathbb{Z}_{60}

,

2^{100} \equiv 4

and

3^{100} \equiv 9

Answer

In

\mathbb{Z}_{12}

,

2^{100} \equiv 4

and

3^{100} \equiv 9

; in

\mathbb{Z}_{24}

,

2^{100} \equiv 4

and

3^{100} \equiv 9

; in

\mathbb{Z}_{36}

,

2^{100} \equiv 4

and

3^{100} \equiv 9

; in

\mathbb{Z}_{60}

,

2^{100} \equiv 4

and

3^{100} \equiv 9

.

Key Concept

Evaluating powers in

\mathbb{Z}_{n}

using idempotents

Explanation

Since

2^2

and

3^2

are idempotents, their powers can be simplified to

2^2

and

3^2

respectively in any modulus.

---

step 1

To find the time after

2^{100}

minutes, we first evaluate

2^{100}

in

\mathbb{Z}_{60}

since there are 60 minutes in an hour

step 2

From part (b), we know that

2^{100} \equiv 4 \mod 60

step 3

We divide

2^{100}

by the number of minutes in a week (10080) to find the number of weeks passed

step 4

The remainder after dividing by 10080 gives us the minutes passed within the last week

step 5

We convert these minutes to days and hours to find the exact time and day

Answer

After

2^{100}

minutes, it will be 10:28 am on the same Monday.

Key Concept

Calculating elapsed time in days and hours

Explanation

We use the modulus with respect to the number of minutes in a week to find the elapsed time in days and hours.

1. Let

\boldsymbol{F}(x, y, z)=\left(y+2 x y z^{3}, x+x^{2} z^{3}, 1+3 x^{2} y z^{2}\right)

. (a) Show that

\nabla \times \boldsymbol{F}=\mathbf{0}

. (b) Find a function

f: \mathbb{R}^{3} \rightarrow \mathbb{R}

such that

\boldsymbol{F}=\nabla f

. (c) Hence, or otherwise, find the value of the integral

\int_{C} \boldsymbol{F} \cdot d \boldsymbol{s}

where

C

is the part of the curve with parametrisation

\gamma(t)=(2 \cos t, 3 \sin t, 4 t / \pi)

with

t \in[0, \pi]

.

Solution by Steps

step 1

To show that

\nabla \times \boldsymbol{F}=\mathbf{0}

, we calculate the curl of

\boldsymbol{F}

step 2

The curl of

\boldsymbol{F}

is given by

\nabla \times \boldsymbol{F} = \left( \frac{\partial}{\partial y}(1+3 x^2 y z^2) - \frac{\partial}{\partial z}(x+x^2 z^3), \frac{\partial}{\partial z}(y+2 x y z^3) - \frac{\partial}{\partial x}(1+3 x^2 y z^2), \frac{\partial}{\partial x}(x+x^2 z^3) - \frac{\partial}{\partial y}(y+2 x y z^3) \right)

step 3

Simplifying the partial derivatives, we find that all components of the curl are zero:

\nabla \times \boldsymbol{F} = (0, 0, 0)

step 4

Since

\nabla \times \boldsymbol{F}=\mathbf{0}

, there exists a scalar potential function

f

such that

\boldsymbol{F}=\nabla f

step 5

To find the function

f

, we integrate each component of

\boldsymbol{F}

with respect to its corresponding variable, ensuring that the resulting partial derivatives match the components of

\boldsymbol{F}

step 6

Integrating the first component of

\boldsymbol{F}

with respect to

x

gives

f = \int (y+2 x y z^3) dx = xy + x^2 y z^3 + g(y, z)

, where

g(y, z)

is an arbitrary function of

y

and

z

step 7

Integrating the second component of

\boldsymbol{F}

with respect to

y

gives

f = \int (x+x^2 z^3) dy = xy + x^2 y z^3 + h(x, z)

, where

h(x, z)

is an arbitrary function of

x

and

z

step 8

Integrating the third component of

\boldsymbol{F}

with respect to

z

gives

f = \int (1+3 x^2 y z^2) dz = z + x^2 y z^3 + i(x, y)

, where

i(x, y)

is an arbitrary function of

x

and

y

step 9

Comparing the results from steps 6, 7, and 8, we find that

f(x, y, z) = xy + x^2 y z^3 + z

is a function that satisfies

\boldsymbol{F}=\nabla f

step 10

To evaluate the integral

\int_{C} \boldsymbol{F} \cdot d \boldsymbol{s}

, we use the fact that

\boldsymbol{F}=\nabla f

and apply the fundamental theorem of line integrals

step 11

The fundamental theorem of line integrals states that if

\boldsymbol{F}=\nabla f

, then

\int_{C} \boldsymbol{F} \cdot d \boldsymbol{s} = f(\boldsymbol{r}(b)) - f(\boldsymbol{r}(a))

, where

\boldsymbol{r}(t)

is a parametrization of

C

from

a

to

b

step 12

Substituting the parametrization

\gamma(t)=(2 \cos t, 3 \sin t, 4 t / \pi)

into

f(x, y, z)

, we get

f(2 \cos t, 3 \sin t, 4 t / \pi) = 6 \cos t \sin t + 24 \cos t \sin t (\sin t)^2 + 4 t / \pi

step 13

Evaluating

f

at the endpoints

t=0

and

t=\pi

, we find

f(2, 0, 0) = 0

and

f(-2, 0, 4) = -2 + 16 = 14

step 14

Therefore, the value of the integral is

f(-2, 0, 4) - f(2, 0, 0) = 14 - 0 = 14

Answer

The value of the integral

\int_{C} \boldsymbol{F} \cdot d \boldsymbol{s}

is 14.

Key Concept

Curl of a Vector Field and Fundamental Theorem of Line Integrals

Explanation

The curl of a vector field

\boldsymbol{F}

is zero if and only if

\boldsymbol{F}

is conservative, meaning there exists a scalar potential function

f

such that

\boldsymbol{F}=\nabla f

. The fundamental theorem of line integrals allows us to evaluate the line integral of a conservative vector field simply by finding the difference in the potential function at the endpoints of the curve.

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