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Would it be an erroneous proof if I only prove one case is true and then claim everything else to be true. For example, prove that $2x-4 > 0$ for all $x|x\in \mathbb{Z},$ $x>2.$ Would this proof be erroneous? The smallest value is $3.$ When $x=3, 2(3)-4>0.$ Therefore this statement is correct.

Joe
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    $2.5$ is smaller than $3$, so $3$ is not the smallest. But even if you only care about integers, you need to say more before you can claim a complete proof. For example, you may argue that $2x-4$ is an increasing function, so the inequality for $x=3$, implies the inequality for every bigger $x$. – markvs Sep 18 '20 at 02:15
  • You also need to state (the obvious) the function is increasing function of x. Also you need to handle 3 > x > 2. – herb steinberg Sep 18 '20 at 02:17
  • There are cases where you can say "without loss of generality" or "other cases are similar" but in this case you need more justification. Certainly it could be made to work but you would need justification. – Luke Sep 18 '20 at 02:17

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Yes, it will be an erroneous proof, because you need to recall that disproving something can be done by only a counter example, while we can prove it by showing that all the cases are true.

As for your proof, we can proceed by letting a function $f(x)=2x-4$. Then, $$f(t+1)-f(t)=2t+2-4-2t-4=2>0$$ Thus, the function is increasing (and in fact, is an increasing AP with common difference $2$).

Now, note that $f(2)=4-4=0$. Since $f$ is increasing, for any $m>2$, $f(m)$ will be always greater than $0$. Thus, we conclude that $2x-4>0$.

Hope this helps :)

ultralegend5385
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