# Abstract Algebra any help is appreciated

• lostinmath08
In summary, on the set of real numbers R, the operation (x,y) -> 2(x,y) - xy - 2 is defined. The neutral element of this operation is 1. However, it has not been proven that a neutral element exists, only that if it does, it must be 1.
lostinmath08
1. On the set of real numbers, R the following operation is defined:
*RxR implies (arrow) R, (x,y) implies (arrow) x*y=2(x+y)-xy-2
Find the neutral element of this operation.

3. since we know x*e=x, e*x=x, so i attempted:
using e as y, because it would just mean y.

so
2(x+e)-xe-2= 2x+2e-xe-2 = 2x+e(2-x)-2 = 2

so wouldn't my e be 2?

i think i am right, but am not too sure, any input would be appreciated!

Try that again. You should solve e*y=y, or 2*(e+y)-ey-2=y. Solve that for e.

i know where i made my mistake the first time, so e=1...im thinking.
this is how i got to my second conclusion:
2(e+y)-ey-2=y
2e+2y-ey-2=y
2e-ey+2y-2=y
e(2-y)=y-2y+2
e(2-y)=(2-y)
e=(2-y)/(2-y)
e=1

so my neutral element would be 1.
thanks for the help!

Since this is not symbolic logic, that $\rightarrow$ is not "implies". It simply indicates that the function if "from" R x R "to" R and changes (x, y) into 2(x+ y)- xy- 2.

lostinmath08 said:
e(2-y)=(2-y)
e=(2-y)/(2-y)
This step is sketchy -- this step is only permissible if 2-y is known to be (globally) nonzero. Some methods for getting around this problem by handling 2-y=0 as a separate case, making an explicit assumption that 2-y is not zero, managing to translate the problem appropriately into univariate polynomials, or my favorite: factoring rather than dividing.

e=1

so my neutral element would be 1.
thanks for the help!
That's not what you've shown! You've shown that if there is a neutral element, it has to be 1. You have not actually shown there exists a neutral element. Fortunately, it's easy to prove that 1 is, in fact, a neutral element.

Last edited:
thank you!

## 1. What is Abstract Algebra?

Abstract Algebra is a branch of mathematics that deals with the study of algebraic structures, such as groups, rings, and fields. It focuses on the general properties and structures of these algebraic objects rather than specific numbers or equations.

## 2. Why is Abstract Algebra important?

Abstract Algebra is important because it provides a foundation for other areas of mathematics, such as number theory, geometry, and topology. It also has applications in fields such as physics, computer science, and cryptography.

## 3. What are some key concepts in Abstract Algebra?

Some key concepts in Abstract Algebra include groups, which are sets with a binary operation that follows certain rules, rings, which are sets with two binary operations, and fields, which are sets with two binary operations that satisfy additional conditions.

## 4. What are some real-life applications of Abstract Algebra?

Abstract Algebra has many real-life applications, such as in coding theory, where it is used to design error-correcting codes, and in cryptography, where it is used to create secure encryption algorithms. It also has applications in physics, chemistry, and economics.

## 5. Is Abstract Algebra difficult to learn?

Abstract Algebra can be challenging to learn, as it deals with abstract concepts and requires a strong foundation in algebra and mathematical reasoning. However, with dedication and practice, it can be a rewarding and fascinating subject to study.

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