# A bit more thinking

1. Aug 5, 2008

### 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?

2. Aug 5, 2008

### HallsofIvy

Staff Emeritus
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.)

Last edited: Aug 6, 2008
3. Aug 5, 2008

### d_leet

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

4. Aug 5, 2008

### uman

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

5. Aug 6, 2008

### HallsofIvy

Staff Emeritus
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.

6. Aug 11, 2008

### 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 $$\frac{-a}{a+1}$$.

7. Aug 11, 2008

### 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)).

8. Aug 11, 2008

### Alex6200

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.

9. Aug 11, 2008

### d_leet

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.

10. Aug 11, 2008

### Alex6200

Yeah, solving it algebraically seems like a good approach.

11. Aug 11, 2008

### CRGreathouse

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

12. Aug 12, 2008

### Mute

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.