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2 years ago

Prove that (I ,+) is an abelian group where I = Set of integers

Mathematics

1 answer:

scoundrel [369]2 years ago

6 0

Step-by-step explanation:

First we recall the relevant definitions and properties:

An even integer is an integer that is a multiple of 2, that is, an integer that can be written as 2k2k where kk is also an integer.

An abelian group is a set with an operation that is closed in that set, is associative, has an identity element, has inverses, and is commutative.

Addition is already associative and commutative over the set of all integers, and has an identity 00 and an inverse −n−n for each integer nn.

Oh, and multiplication of integers distributes over addition (this is important because we’re dealing with multiples of 2 but also with addition. The distributive property is how multiplication relates to addition).

This means we have to show a few things:

Addition is closed over the even integers. This holds due to the distributive property: if you have even integers 2k2k and 2m2m, then 2k+2m=2(k+m)2k+2m=2(k+m) is also an even integer. The odd integers fail this property: for example, 11 is odd but 1+1=21+1=2, which is not odd.

Addition is associative over the even integers. This holds because addition is already associative over the set of all integers: 2k+(2m+2j)=(2k+2m)+2j2k+(2m+2j)=(2k+2m)+2j. The odd integers do satisfy associativity, since they’re also a subset of the integers.

Addition has an identity element over the even integers. Since we already know that 00 is an identity for the set of all integers and 00 is even, this shows that we have an identity for the even integers: 2k+0=2k2k+0=2k. This doesn’t hold for the set of odd integers, because if nn and kk are odd integers and n+k=nn+k=n then k=0k=0, a contradiction since 00 is not odd.

Addition has inverses over the even integers. We already know that integers have inverses, and if 2k2k is an even integer then −k−k is the inverse of kk, so that 2k+2(−k)=2(k+(−k))=2(0)=02k+2(−k)=2(k+(−k))=2(0)=0. This means the even integer 2(−k)2(−k) is the inverse of 2k2k. The odd numbers do satisfy this property, since they’re also a subset of the integers.

Addition is commutative over the even integers. This holds because addition is already commutative over the set of all integers: 2k

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Write an equation for the description.

cricket20 [7]

I think it would be “ 130 + m = 170 “ I could be wrong tho

6 0

3 years ago

Read 2 more answers

Consider the integral Integral from 0 to 1 e Superscript 6 x Baseline dx with nequals 25 . a. Find the trapezoid rule approximat

photoshop1234 [79]

Answer:

With n = 25, $\int_{0}^{1}e^{6 x}\ dx \approx 67.3930999748549$

With n = 50, $\int_{0}^{1}e^{6 x}\ dx \approx 67.1519320308594$

b. $\int_{0}^{1}e^{6 x}\ dx \approx 67.0715427161943$

The absolute error in the trapezoid rule is 0.08047

The absolute error in the Simpson's rule is 0.00008

Step-by-step explanation:

a. To approximate the integral $\int_{0}^{1}e^{6 x}\ dx$ using n = 25 with the trapezoid rule you must:

The trapezoidal rule states that

$\int_{a}^{b}f(x)dx\approx\frac{\Delta{x}}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)\right)$

where $\Delta{x}=\frac{b-a}{n}$

We have that a = 0, b = 1, n = 25.

Therefore,

$\Delta{x}=\frac{1-0}{25}=\frac{1}{25}$

We need to divide the interval [0,1] into n = 25 sub-intervals of length $\Delta{x}=\frac{1}{25}$ , with the following endpoints:

$a=0, \frac{1}{25}, \frac{2}{25},...,\frac{23}{25}, \frac{24}{25}, 1=b$

Now, we just evaluate the function at these endpoints:

$f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1$

$2f\left(x_{1}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281$

$2f\left(x_{2}\right)=2f\left(\frac{2}{25}\right)=2 e^{\frac{12}{25}}=3.23214880438579$

...

$2f\left(x_{24}\right)=2f\left(\frac{24}{25}\right)=2 e^{\frac{144}{25}}=634.696657835701$

$f\left(x_{25}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735$

Applying the trapezoid rule formula we get

$\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{50}(1+2.54249830064281+3.23214880438579+...+634.696657835701+403.428793492735)\approx 67.3930999748549$

To approximate the integral $\int_{0}^{1}e^{6 x}\ dx$ using n = 50 with the trapezoid rule you must:

We have that a = 0, b = 1, n = 50.

Therefore,

$\Delta{x}=\frac{1-0}{50}=\frac{1}{50}$

We need to divide the interval [0,1] into n = 50 sub-intervals of length $\Delta{x}=\frac{1}{50}$ , with the following endpoints:

$a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b$

Now, we just evaluate the function at these endpoints:

$f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1$

$2f\left(x_{1}\right)=2f\left(\frac{1}{50}\right)=2 e^{\frac{3}{25}}=2.25499370315875$

$2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281$

...

$2f\left(x_{49}\right)=2f\left(\frac{49}{50}\right)=2 e^{\frac{147}{25}}=715.618483417705$

$f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735$

Applying the trapezoid rule formula we get

$\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{100}(1+2.25499370315875+2.54249830064281+...+715.618483417705+403.428793492735) \approx 67.1519320308594$

b. To approximate the integral $\int_{0}^{1}e^{6 x}\ dx$ using 2n with the Simpson's rule you must:

The Simpson's rule states that

$\int_{a}^{b}f(x)dx\approx \\\frac{\Delta{x}}{3}\left(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right)$

where $\Delta{x}=\frac{b-a}{n}$

We have that a = 0, b = 1, n = 50

Therefore,

$\Delta{x}=\frac{1-0}{50}=\frac{1}{50}$

We need to divide the interval [0,1] into n = 50 sub-intervals of length $\Delta{x}=\frac{1}{50}$ , with the following endpoints:

$a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b$

Now, we just evaluate the function at these endpoints:

$f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1$

$4f\left(x_{1}\right)=4f\left(\frac{1}{50}\right)=4 e^{\frac{3}{25}}=4.5099874063175$

$2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281$

...

$4f\left(x_{49}\right)=4f\left(\frac{49}{50}\right)=4 e^{\frac{147}{25}}=1431.23696683541$

$f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735$

Applying the Simpson's rule formula we get

$\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{150}(1+4.5099874063175+2.54249830064281+...+1431.23696683541+403.428793492735) \approx 67.0715427161943$

c. If B is our estimate of some quantity having an actual value of A, then the absolute error is given by $|A-B|$

The absolute error in the trapezoid rule is

The calculated value is

$\int _0^1e^{6\:x}\:dx=\frac{e^6-1}{6} \approx 67.0714655821225$

and our estimate is 67.1519320308594

Thus, the absolute error is given by

$|67.0714655821225-67.1519320308594|=0.08047$

The absolute error in the Simpson's rule is

$|67.0714655821225-67.0715427161943|=0.00008$

6 0

3 years ago

Solve for x.<br>3 - |8x-6I = 3<br><br>x= or x=

Ket [755]

Answer:

¾ = x

Step-by-step explanation:

Since 3 - 0 gives us the 3, we do this:

$0 = -|8x - 6|$

0 = -8x + 6 >> Distributive Property

-6 - 6

___________

-6 = -8x

__ ___

-8 -8

¾ = x

I am joyous to assist you anytime.

7 0

3 years ago

I know this one is hard but its hard for me. PLEASE HELPP

xenn [34]

Answer:

G. <3 and <5 hope this helps

5 0

2 years ago

Read 2 more answers

Fill in the integers -10, -8, -7, -6, -5, -4, -3, -2 and 0 into a magic square so that every row, column, and diagonal has a sum

ioda

-2 -10 -3

-7 0 -8

-6 -5 -4

That's the square, it was a fun puzzle :)

6 0

2 years ago