Math Help - Is F(x)=X*((x+1)/(x+1)) equal to G(x) = X*1?

1. Is F(x)=X*((x+1)/(x+1)) equal to G(x) = X*1?

I have been informed that F(x)=X*((X+1)/(X+1)) is not equal to Y=X since the former is undefined at X=-1

So, F(x)=X ((X+1)/(X+1)) is not equal to Y=(1)X, am I then saying that (X+1)/(X+1) is not equal to (1)?

At this time to me this seems to imperil some pretty basic operations, for example take combining fractions.

D(x) = (1/x) + 2/(1+X) = 2

I have been taught that we combine fractions by multiplying each fraction by 1, in this case (X/X) and (X+1)/(X+1). Does this procedure increase the number of points at which the function D(x) is undefined? Or, wouldn't D(x) become some new function?

In addition, any function can be multiplied by one and remain unchanged. So P(x) can be said to have an infinite string of 1's being multiplied to it. If, (X+1)/(X+1) = 1, is true and F(x) is also undefined at -1 then couldn't it be said that a function U(x) = (X+2)/(X+2) exits and that F(x)*U(x) is equal to one and undefined at -2 and -1. Repeating this procedure for all, lets say counting numbers, would mean that any function is undefined for all counting numbers? This makes me think that no function can exist as it would be undefined at all possible numbers.

This whole thing seems clearly wrong to me.

I couldn't think of a better spot for this question within the forum structure. As far as I know this question of mine doesn't relate to any specific class, please feel free to move it as is appropriate.

Also I am interested in both narrow and broad corrections/comments. If this is more then just a simple error (as unlikely as that is) and relates to something complicated please let me know what that complicated thing is called.

2. Re: Is F(x)=X*((x+1)/(x+1)) equal to G(x) = X*1?

Originally Posted by bkbowser
I have been informed that F(x)=X*((X+1)/(X+1)) is not equal to Y=X since the former is undefined at X=-1
Yes. On one level, functions are defined as a subset of the cartesian product of a domain set and a codomain set. In order for two functions to be equal you need to show (among other conditions) set equivalence. In your example, $(-1,-1) \not\in F$, yet $(-1,-1) \in G$, so $F\neq G$.
Originally Posted by bkbowser
am I then saying that (X+1)/(X+1) is not equal to (1)?
Your confusion comes from glossing over the understood conventions of algebra. When one uses an algebraic variable in an expression (such as $x$), this acts as a placeholder for members of a field (usually the field of real or complex numbers). When one evaluates the truth of a statement involving variables, it is understood that the statement should be true for all substitutions of the variables over the understood field. For example, the equation $x+1=x+1$ is true for all real numbers substituted for $x$.

On the other hand, the equation $\frac{x+1}{x+1}=1$ does not hold for all real numbers $x$ (in particular, the left hand side is undefined when $x=-1$). Often, when the points where a statement fails are easy to determine, we note the exceptions and continue using the statement. For example, we could say that $\frac{x+1}{x+1}=1$ is true provided $x\neq-1$. Any further algebraic results using the above statement would continue to have the exception noted.
Originally Posted by bkbowser
At this time to me this seems to imperial some pretty basic operations, for example take combining fractions.

D(x) = (1/x) + 2/(1+X) = 2
I think you did something wrong here. But just for the sake of argument, for the rest of the post I will assume that you wrote $D(x)=\tfrac{1}{x}+\tfrac{2}{1+x}-\tfrac{1}{x(1+x)}=\tfrac{3}{1+x}$
Originally Posted by bkbowser
I have been taught that we combine fractions by multiplying each fraction by 1, in this case (X/X) and (X+1)/(X+1).
Remember that $\tfrac{x}{x}=1$ provided $x\neq0$ and $\tfrac{x+1}{x+1}=1$ provided $x\neq-1$.
Originally Posted by bkbowser
Does this procedure increase the number of points at which the function D(x) is undefined?
Yes, exactly. $D(x)$ would not be defined if $x=0$ or $x=-1$.
Originally Posted by bkbowser
Or, wouldn't D(x) become some new function?
Yes, $D(x)$ is a different function from the function $\tfrac{3}{1+x}$ (the right hand side is already undefined at $x=-1$). We might write $D(x)=\tfrac{3}{1+x}$ provided $x\neq0$. Another way of is $D(x)=\begin{cases}\frac{3}{1+x}, & \text{if }x\neq0\\ \text{undefined}, & \text{otherwise}\end{cases}$.

I think from here, you should be able to answer the rest of your questions.