# Thread: When showing your workings how often should you state when x != 0

1. ## When showing your workings how often should you state when x != 0

It may have multiple terms of a denominator of x. Should you declare x != 0 on every step of your simplification process. Or only write it on your final result? What is considered the best practice ? I currently write it on every step of my calculations just to avoid losing a mark (i have no idea how the examiner might feel about it). So i play it safe.

But in general - what is a the common practice?

2. ## Re: When showing your workings how often should you state when x != 0

Whenever you end up dividing by any expression that can potentially equal 0 you have to state that you are assuming it doesn't.

If your expression has multiple factors any of which might be 0 you simply have to state that you are assuming the entire expression is non-zero.

You don't have to specify that you are assuming each individual factor is non-zero. That is obvious given the entire expression is non-zero.

Once you have stated that assumption for any particular expression you don't have to repeat it, but should you end up dividing by a new expression you do have to state that you are assuming this new expression is non-zero.

3. ## Re: When showing your workings how often should you state when x != 0

In addition to what Romsek said, if you do assume that an expression is non-zero as a part of your calculation, you should usually then consider what happens if that quantity were zero.

For example:
\begin{align*}a^2 &= b^2 \\
a^2 - b^2 &= 0 \\
(a+b)(a-b) &= 0 \\
a + b &= 0 & ( a - b \ne 0) \\
\implies a &= -b \end{align*}
but if $a-b = 0$, we get $a=b$. And thus when $a^2 = b^2$ we have two possibilities
1. $a = -b$ (if $a - b \ne 0$); or
2. $a=b$

Another example:
\begin{align*} \text{Let} \quad S &= 1+ x + x^2 + \ldots + x^n \\
\text{then} \quad xS &= \phantom{1+{}} x + x^2 + \ldots + x^n + x^{n+1} \\
\text{so} \quad S-xS = (1-x)S &= 1 - 0 - 0\phantom{{}^2} - \ldots - 0\phantom{{}^n} - x^{n+1} \\
\text{and} \quad S &= \frac{1-x^{n+1}}{1-x} & (x \ne 1)\end{align*}
When $x=1$ we have $$S = 1 + (1) + (1)^2 + \ldots + (1)^n = \underbrace{1 + 1 + 1 + \ldots + 1}_{\text{(n+1) terms}} = n+1$$