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Currently trying to explain some maths to a friend.

He has taken a statement $x^2 + 4 > 2x$ and tried to prove this is true for all $x$.

His proof is $x^2+4>2x \Rightarrow x^2-2x + 4 > 0 \Rightarrow (x-1)^2 + 3 > 0$ which is true so the original statement is true.

However this starts at the wrong place and the implication goes in the wrong direction. So I think it’s wrong and I can’t seem to convince him of this or find some basic examples to illustrate the point that statement X $\rightarrow$ true statement doesn’t mean that X is true....

So can anyone explain to me why it’s wrong using some basic counterexamples perhaps so I can have the knowledge to explain why it is wrong...

Thanks

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    $\begingroup$ The proof should demonstrate that the last statement implies the first and that the last is true, therefore concluding that the first will be true. However, those implication arrows are actualy are equivalences, so that is done. $\endgroup$ – Graham Kemp Nov 10 '18 at 12:53
  • $\begingroup$ @GrahamKemp Yes I know, but they’re left out so it’s not correct in proving the statement. I am just looking for example that if statement X gives a true statement then X is not true $\endgroup$ – PhysicsMathsLove Nov 10 '18 at 12:54
  • $\begingroup$ Indeed, $X\to \top$ does not entail that $X$ is true. However, those arrows should be flipped, and $X\gets \top$ does entail that $X$ is true. $\endgroup$ – Graham Kemp Nov 10 '18 at 13:01
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Your friend is correct, the subtlety is that all his steps are reversible, so a clear way to put it is as: $$ x²+4>2 \iff x²-2x+4>0 \iff (x-1)²+3>0 $$ This way the truthiness of the last statement implies the same for the first. But you are correct to be cautious, a case where things would go wrong is with squares. For example: $$ x=1 \Rightarrow x² = 1 \Rightarrow x=1~\text{or}~x =-1 $$ The last sentence is true if $x=-1$, but the first would be false.

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  • $\begingroup$ Thank you. Yes I know that his steps are reversible and you can just place iff symbols, but without them, it is wrong right? It's like if you want to prove $a^2 + b^2 > 2ab$, you don't go $a^2 - 2ab + b^2 \geq 0$, which gives $(a-b)^2 \geq 0$, so statement is true, you either put iff symbols there or start at $(a-b)^2 \geq 0$? $\endgroup$ – PhysicsMathsLove Nov 10 '18 at 13:00
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    $\begingroup$ Exactly. The proof as given was incorrect, but easily correctable. @PhysicsMathLover $\endgroup$ – Graham Kemp Nov 10 '18 at 13:11
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Yes the fact is that the following chain

$$x^2+4>2x \Rightarrow x^2-2x + 4 > 0 \Rightarrow (x-1)^2 + 3 > 0$$

is true and it can be useful to guess a way for the proof which requires the following

$$(x-1)^2 + 3 > 0 \Rightarrow x^2-2x + 4 > 0 \Rightarrow x^2+4>2x $$

or in a more direct way we can use

$$x^2+4>2x \iff x^2-2x + 4 > 0 \iff (x-1)^2 + 3 > 0$$

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