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For direct proof, if an exception appears, like for example a(a-1) is always divisible by 2 except for the value 1. Do we mention the exception or we need to find an alternative way of explaining to not mention the exception? I am new to discrete maths so pardon my lack of knowledge.

MikeyYo Ho
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    When $a=1$ it is $0$, which is still divisible by $2$. – Randall Jul 29 '20 at 01:59
  • Could you give an example of an alternative way of explaining that does not mention the exception? – Adina Goldberg Jul 29 '20 at 02:02
  • Well, the alternative way I was thinking of is by subbing values directly in the formula. Like let a = 2 in order to show the proof. But that is incorrect is it? Sorry, I thought 0 being divided by 2 is like an error because output is zero. – MikeyYo Ho Jul 29 '20 at 02:07
  • 0 divided by 2 is equal to 0. It is not an error. 2 divided by 0 would be an error. – Ted Jul 29 '20 at 02:32
  • Thank you for the clarification! Although I never really see exception in direct proof, so I am not sure if it is not supposed to be written. – MikeyYo Ho Jul 29 '20 at 02:36

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What is important here is the set of values for which you are proving the statement. For example, we can prove that: $$\forall N\in\Bbb N;\ \sum_{k=1}^{N}\frac{1}{k(k+1)}=\frac{N}{N+1}$$ But this statement is complete nonsense for any $N\in \Bbb R/\{\Bbb N\}$

Let's assume for argument's sake that $a=1$ was a counterexample to some statement $A$. We would then say, after proving the rest of $\Bbb N$ (or $\Bbb Z, \Bbb R$, etc), that $A$ is true for all $a\in \Bbb N/\{1\}$.

Rhys Hughes
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