I think I have a pretty good understanding of implication and equivalence (I also found this question), but there are some things I am unsure about.

First of all, in maths class in high school, when for example given a quadratic function $f(x)$ with the derivative $f'(x)=2ax+b$, to compute the extremum of $f$ we would write something like

$$ f'(x)=0 \\ \Downarrow \\ 2ax+b=0 \\ \Updownarrow \\ x=-\frac{b}{2a} $$

formatting notwithstanding.

I understand why the second implication is an equivalence, but I'm not so sure why the first is not. (I also think I have seen Sal Khan on Khanacademy write a double arrow in cases like this, though I'm not sure.) I guess my confusion springs from the following, considering the growth of a linear function $f(x)=ax+b$ in the interval $\Delta x$:

$$ f(x+\Delta x)=a(x+\Delta x)+b=ax+b+a\Delta x=f(x)+a\Delta x $$

(By the way, this was printed in my sophomore textbook.) Here, in the third step, the function $f(x)$ is inserted into the expression. Then, if these are equal (which they obviously are), why is the first implication above not an equivalence?

Secondly, I previously asked this question, but later stumbled upon these notes. I'm a bit confused about the use of arrows on page 29. First, this:

Here, I understand the use of the arrows to mean, that each expression on the right sides are equivalent to each other, the last expression being equivalent to the first one on the right, which in turn is equivalent to the left hand expression. Then, this:

Here, I assume the three right hand expressions are equivalent? In that case, what I don't understand is the use of single arrows. I assume the formatting could imply that the first expression implies the three right hand expressions and that the right hand expressions are not explicitly stated to be equivalent

About the question I asked, I was also wondering if this would be an acceptable way of formatting for instance the solving of an equation (like the quadratic above)?

I would very much appreciate if someone could shed some light on this. Thanks.

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    $\begingroup$ I only skimmed through the question, but in all instances it seems to comes down to the following. It's not that they weren't equivalences, it's that the given implications suffice for the purposes of whatever is being done. $\endgroup$
    – Git Gud
    Oct 14, 2014 at 14:24
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    $\begingroup$ If we start from $f′(x)=2ax+b$, we are saying that the LHS and the RHS are two "names" of the same object. Thus the function "named" $f'(x)$ will get the value $0$ exactly for those values of $x$ for which the function "named" $2ax+b$ will, because they are the "same" function. $\endgroup$ Oct 14, 2014 at 14:26
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    $\begingroup$ From context I would say $A\iff B$ is, by definition, $A\implies B\land B\implies A$. In the first instance an equivalence is correct for the reason Mauro stated. And I didn't check, but it seems to me like all the $\implies$'s here can be replaced by $\iff$'s. $\endgroup$
    – Git Gud
    Oct 14, 2014 at 14:33
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    $\begingroup$ @GitGud "It's not that they weren't equivalences, it's that the given implications suffice for the purposes of whatever is being done." I agree completely. $\endgroup$ Oct 14, 2014 at 14:34
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    $\begingroup$ YES. We can think at "Mauro is the father of Laura" as "Mauro = the father of Laura"; then "Mauro is 55 years old" is true iff "the father of Laura is 55 years old". We use in math the symbol "=" in two ways : (i) as identity : $f'(x)=2ax+b$ is a relation between two names of the same thing. In this case we can freely substitute one name for the other. (ii) as an equation : $f'(x)=0$ does not mean that $f'(x)$ is identically $0$; it means that we are "asking for" one (or more) values of $x$, call it $x_0$ for which the equation $f'(x_0)=0$is satisfied. 1/2 $\endgroup$ Oct 14, 2014 at 14:42

1 Answer 1


Your doubts can be explained by the following theorem: $$\left(A\Leftrightarrow B\right) \Rightarrow \left(A\Rightarrow B\right)$$ Same holds for the other direction, of course.

Practically spoken, this means that if two statements are equivalent, it is not a false statement to use an implication there.

The reason why most people do this is because in many proofs, you just have to show that some implication is true, which is why you would structure your proof like the pattern $$A\Rightarrow B\Rightarrow \dots\Rightarrow X$$

But in such a case, it is rather consequent to just use the implication-part of any equivalence, because 1) It would be correct (see above) and 2) You don't have to concern the question if a certain statement is an equivalence or not, it just comes in handy.

But the main thing to recognize here is, it just doesn't matter -- as long as you are not trying to prove the $"\Leftarrow"$ direction.

BTW the Proof (although quite trivial):

$$\left(A\Leftrightarrow B\right) \Leftrightarrow \left( \left(A\Rightarrow B\right) \land \left(A\Leftarrow B\right) \right) \Rightarrow \left(A\Rightarrow B\right)$$


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