Finding the Identity and Inverse of the Operation *

[roam]

"Define a * b, for a,b \in R by a * b = ab+a+b. What we just have done is defined a new operation *, in terms of the well known ones multiplication and addition."

Is there a real number which is the identity under the operation * ?
Lets call this number e. A number for which a * e = e * a for all a \in R?

Furtheremore, does every a \in R have an inverse under this operation? ...if so, what could it be?

 

[HallsofIvy]

Looks an awful lot like a homework problem to me!

To answer the first question, you are looking for a number e such that a*e= ae+ a+ e= a. That means ae+ e= 0 so ae= e for all a. There is no real number for which that is true so this operation has no identity.

(By the way "a*e= e*a" is true but does NOT define the identity. for example that is true for any member of a commutative group.)

 

[d_leet]

Looks an awful lot like a homework problem to me!

To answewr the first question, you are looking for a number e such that a*e= ae+ a+ e= a. That means ae+ e= 0 so ae= e for all a. There is no real number for which that is true so this operation has no identity.

Err... Doesn't zero have this property?

 

[uman]

Yep. a * 0 = a0 + a + 0 = a.

 

[HallsofIvy]

Oh, bother!

Okay, now that it has an identity, you need to find inverses for this operation. If b is a's inverse, then ab+ a+ b= 0. Solve for b.

 

[roam]

I think so too. But if e is the identity, we should have a*e = ae+a+e=a and e*a=ea+e+a=a
=> e(a+1)=0 and e = 0 And yes 0 is the only solutions that works for every a \in R

In order to find the inverse we want for every a, a real number b for which a*b=ab+a+b=e=0

Then b(a+1)=0 => b= \frac{-a}{a+1} and we DO have an inverse under this operation for every real number except -1, by letting the inverse of a be the number \frac{-a}{a+1}.

 

[gel]

Take any invertible function f:R->R and you can define an operation a*b = f^-1( f(a) f(b) ). The identity is f^-1(1). The inverse of a is f^-1(1/f(a)).

 

[Alex6200]

Yep, the answer is 0.

But just try different numbers to find an identity. Eventually you'll find it. Or you can do it algebraically:

a * b = ab + a + b = a (true for identity, since a * b = a)

ab + a + b = a

ab + b = 0;

(a+1)b = 0

For a = -1, any b is an identity. For a \neq -1, b = 0.

 

[d_leet]

Yep, the answer is 0.

But just try different numbers to find an identity. Eventually you'll find it. Or you can do it algebraically:

a * b = ab + a + b = a (true for identity, since a * b = a)

ab + a + b = a

ab + b = 0;

(a+1)b = 0

For a = -1, any b is an identity. For a \neq -1, b = 0.

Just trying numbers is not a good method, especially when there are uncountably many to choose from. And b=0 works even when a=-1.

 

[Alex6200]

Just trying numbers is not a good method, especially when there are uncountably many to choose from. And b=0 works even when a s1.

Yeah, solving it algebraically seems like a good approach.

 

[CRGreathouse]

Just trying numbers is not a good method, especially when there are uncountably many to choose from. And b=0 works even when a=-1.

I think the definable numbers would suffice, and they're countable. So it's not too hard after all.

 

[Mute]

Is there a real number which is the identity under the operation * ?
Lets call this number e. A number for which a * e = e * a for all a \in R?

Note that since a and b are real numbers and multiplication and addition of real numbers are commutative, a*b = b*a for all real numbers a and b:

a*b = ab + a + b = ba + b + a = b*a, where in the centre we use the fact that ab = ba for real numbers and a + b = b + a for real numbers.