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**Posts by Reek**

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**Reek**- Replies: 1

We can build 1/33 like this, .030303..... (03 repeats). .030303.... tends to 1/33 .

So,I was wondering this: In the decimal representation, if we start writing the 10 numerals in such a way that the decimal portion never ends and never repeats; then am I getting closer and closer to some irrational number?

**Reek**- Replies: 0

formula for getting the volume of revolution about the line (pi)/2 = θ

when f(θ)=r

is

b

∫ (2/3)*(pi)*(r³)*cos(θ) d(θ)

a

How do we derive at that formula?:|

Hello Bobbym;

here are the particulars

−2x^3 − 15x^2 − 6x + 7 = 0

2x^4 − 11x^3 − 6x^2 + 64x + 32 = 0

Thank u bobbym for your assistance.

**Reek**- Replies: 5

How can we derive the formula that calculates the discriminant of a polynomial?

Could anyone please explain the process to me? I'm in a hurry, please do make it quickly.

(I am saying about a polynomial in x over the set of complex numbers.)

Level of mine:University level

5. Show that the set {-1, 1} is a group under multiplication, but not addition.

For any elements a and b of {-1, 1}, (a+b) is not an element of {-1, 1}. It is enough to show only one rule-break to prove that {-1, 1} is not a group with respect to addition.

We conclude {-1, 1} is not a group with respect to addition.

------------------------------------------------------------------------------------------------------------------------

For any elements a and b of {-1,1}, (a*b) is an element of {1,-1}. The closure law has been followed.

For any a,b,c of {1,-1}; a*(b*c) = (a*b)*c. The associative law has been followed.

For any a of {1,-1} i*a=a, where i is a particular element in {1,-1}.The left identity element i is 1 here.

For any a of {1,-1} the equation x*a=i has a solution known as the left inverse of a.

All these properties are followed by this set that is closed under multiplication.

Therefore, {1,-1} is a group with respect to multiplication.

4. Show that the set {1} with multiplication is a group.

For any elements a and b of {1}, (a*b) is an element of {1}. The closure law has been followed.

For any a,b,c of {1}; a*(b*c) = (a*b)*c. The associative law has been followed.

For any a of {1} i*a=a, where i is a particular element in {1}.The left identity element i is 1 here.

For any a of {1} the equation x*a=i has a solution known as the left inverse of a.1 is the only element in {1} and the left inverse of 1 is 1.

All these properties are followed by this set that is closed under multiplication.

Therefore, {1} is a group with respect to multiplication.

For any elements a and b of {1}, (a+b) is not an element of {1}. 1+1=2. It is enough to show only one rule-break to prove that {1} is not a group with respect to addition.

We conclude {1} is not a group with respect to addition.

2.Show that the set {0} with multiplication is a group.

For any elements a and b of {0}, (a*b) is an element of {0}. The closure law has been followed.

For any a,b,c of {0}; a*(b*c) = (a*b)*c. The associative law has been followed.

For any a of {0} i*a=a, where i is a particular element in {0}.The left identity element i is 0 here.

For any a of {0} the equation x*a=i has a solution known as the left inverse of a.0 is the only element in {0} and the left inverse of 0 is 0.

All these properties are followed by this set that is closed under multiplication.

Therefore, {0} is a group with respect to multiplication.

1. Show that the set {0} with addition is a group.

For any elements a and b of {0}, (a+b) is an element of {0}. The closure law has been followed.

For any a,b,c of {0}; a+(b+c) = (a+b)+c. The associative law has been followed.

For any a of {0} i+a=a, where i is a particular element in {0}.The left identity element i is 0 here.

For any a of {0} the equation x+a=i has a solution known as the left inverse of a.0 is the only element in {0} and the left inverse of 0 is 0.

All these properties are followed by this set that is closed under addition.

Therefore, {0} is a group with respect to addition.

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