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

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

    So say your showing your math workings for simplifying your answer.

    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?
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  2. #2
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    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.
    Last edited by romsek; Feb 18th 2018 at 08:35 PM.
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  3. #3
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    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 $$S = 1 + (1) + (1)^2 + \ldots + (1)^n = \underbrace{1 + 1 + 1 + \ldots + 1}_{\text{$(n+1)$ terms}} = n+1$$
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