The operation $x*y := x+y-1$ gives a group structure on the reals Show that the binary operation * given by the rule $x*y = x+y-1$ defines a group operation on the set of all real numbers.
I know I have to show that it is associative, has an identity element, and has an inverse. 
For associativity 
$(x*y) * z = x * (y *z ) $??? 
So, 
$(x*y) * z = (x + y - 1) * z $
not sure if this is even correct. I really need help with this!
 A: Here's how you go about this:
You have to prove that 
$$
(x*y) * z = x * (y * z)
$$
To do this, you want to convert the left-hand side into something you're familiar with, do the same with the right, and then compare. 
For the left-hand side, we use the rule
$$
u * v = u + v -1
$$
which is just a rewriting of the rule you gave, but with new variable names. 
Apply this to $u = (x*y)$ and $v = z$ to get that 
$$
(x*y) * z = (x*y) + z - 1
$$
Now apply that same rule to $u = x, v = y$ to get an expression with no stars in it. 
Then do the same thing with the right hand side and compare. 
I haven't worked out all the details for you, because that's how you learn. 
But I will observe that the thing you wrote:
$$
(x*y)*z=(x+y−1)*z
$$
is another way to get started (i.e., expand out the inner star first, and then approach the outer one). Both ways work. But you need to apply the rule a second time, and that's why writing it out with other letters (my $u$ and $v$) can help you: if you use the same letters that appear in the thing you're working on, it's hard to do the substitution. You have to write silly things like "Applying this to $x = x*y$ and $y = z$, we get ...", and that really doesn't quite make sense. 
A: To show associativity, you need to write something like this:
$$
(x*y)*z = \cdots\cdots\cdots\cdots = x*(y*z)
$$
and each step should put $\text{“}=\text{''}$ between things you already know are equal. Thus
$$
(x*y)*z = (x*y) + z - 1 = (x+y-1)+z - 1
$$
and
$$
x*(y*z)  = x + (y*z) - 1 = x + (y+z-z) -1
$$
and then, since you can see that the two expressions on the right sides are equal to each other, you can write
\begin{align}
(x*y)*z = (x*y) + z - 1  & = (x+y-1)+z - 1 \\[6pt]
& = x + (y+z-1) -1 = x + (y*z) - 1 = x*(y*z). 
\end{align}
