Rational expressions: cancellation keeps excluded values excluded

In $\frac{x^2-1}{x-1}$ we can cancel $x-1$ to obtain $x+1$ . Does that make $x=1$ allowed? No: the original fraction had no value at $x=1$ , and a shorter form does not change its original conditions. For every other real $x$ , its value does equal $x+1$ .

We need factoring: only common factors can be cancelled, not arbitrary matching pieces of notation.

Record the domain first

A rational expression is a quotient of polynomials. Its domain contains all variable values for which the denominator is nonzero. Here $x-1\ne0$ , so $x\ne1$ . For $1/[x(x+2)]$ , both 0 and −2 are excluded.

Only then do we factor the numerator:

\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1,\qquad x\ne1.

Cancellation uses $(x-1)/(x-1)=1$ , which requires $x-1$ to be nonzero. At $x=1$ we get $0/0$ , not 1: the division has no defined result.

What the open circle means

The horizontal axis shows input $x$ ; the vertical axis shows the fraction’s value $y$ . Each pair $(x,y)$ becomes a point. All allowed pairs lie on $y=x+1$ , but $(1,2)$ is missing. The open circle marks an absent point, not a zero value.

The fraction (x²−1)/(x−1) equals x+1 only for x ≠ 1. The open circle (1,2) is excluded. Scales are uniform; −1 ≤ x ≤ 3 is shown.

At $x=2$ , we get $3/1=3$ ; at $x=0$ , $(-1)/(-1)=1$ . Near one, values are near 2, but proximity does not create a value at the forbidden input.

Terms in a sum cannot be cancelled

$\frac{x+2}{x}$ does not equal 2 in general. At $x=2$ it is 2, but at $x=4$ it is $6/4=1.5$ . The valid split is $\frac{x+2}{x}=1+\frac2x$ , where $x\ne0$ . In the numerator, $x$ is a term, not a separate factor of the entire sum. Rational operations, OpenStax §1.6.

A common denominator and the signs

To add or subtract, use one denominator, changing numerator and denominator together:

\frac1x+\frac1{x+1}=\frac{x+1+x}{x(x+1)}=\frac{2x+1}{x(x+1)}.

Keep the original $x\ne0,-1$ . When subtracting, subtract the whole second numerator: $(x+1)-x=1$ , so the difference is $1/[x(x+1)]$ on the same domain.

For multiplication, multiply numerators and denominators. For example, $\frac{x-1}{x}\cdot\frac{x}{x+2}=\frac{x-1}{x+2}$ , but $x\ne0,-2$ even after $x$ disappears.

Division adds another restriction

A divisor must have a nonzero value. Thus:

\frac{x}{x+1}:\frac{x}{x-1}=\frac{x}{x+1}\cdot\frac{x-1}{x}=\frac{x-1}{x+1}.

The first denominator excludes $-1$ , the second excludes 1, and a nonzero divisor excludes 0. All three restrictions accompany the result. Forgetting the last would permit division by zero, now hidden by cancellation.

Branch · A distinctionFilling in a value is a separate operation.

We can define a new rule

We can define $y=x+1$ for every real $x$ , including one. It extends the original relationship by filling the hole with 2. But it is a new rule with a wider domain, not proof that the old fraction already had that value.

Simplification preserves answers where the original operations made sense. Carry the shorter expression and its original restrictions together.