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Really simple question. Is valid to do this:

$$ 2x + 2 = 1 \iff x = \frac{1 - 2}{2} = -\frac{1}{2} $$

I mean is anything wrong when solved for $x$ to not use $\iff$ symbol more and just use $=$.

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  • $\begingroup$ I would say that this is the most correct way to write the resolution of an equation. $\endgroup$
    – azif00
    Mar 3 at 16:13
  • $\begingroup$ Be careful... Many steps in a solution are not biimplications. For example, in the normal procedures, one also does not write down non-generic alternatives; e.g., $ax+b=0 \iff \left( x = -b/a \vee (a=0 \wedge b=0) \right)$. $\endgroup$ Mar 3 at 16:18
  • $\begingroup$ I think this is a very useful way to distinguish reversible steps. Sometimes, of course, steps are irreversible. $x=y\implies x^2=y^2$ but the reverse implication does not hold generally. In your case, assuming you are working with rational numbers or something akin to that, both implications certainly do hold. $\endgroup$
    – lulu
    Mar 3 at 16:20
  • $\begingroup$ The = sign is used to equate expressions $a=b$, while $\implies$ is used between statements e.g. $a=b \implies c=d$. $\endgroup$
    – Paul
    Mar 3 at 16:27
  • $\begingroup$ Equal between terms, I e names, equivalent between statements $\endgroup$ Mar 3 at 16:52

2 Answers 2

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What you write is correct. I think it would be even better (equally correct and easier to read) with words:

$$ 2x + 2 = 1 $$ implies $$ x = \frac{1-2}{2} = -\frac{1}{2} . $$

Note the period at the end of the sentence - since it is a sentence.

In this case you don't care about the reverse implication, which happens to be true here but isn't always when manipulating expressions.

Replacing the $\iff$ by $=$ would make your argument just plain wrong - in a way all too commonly seen. Read literally it would say $$ 2x + 2 = 1 = \cdots = -\frac{1}{2} . $$

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The equivalence sign is actually pretty smart since many students often write $$ 2x+2=1 \Longrightarrow x=-\dfrac{1}{2} $$ and ignore the fact that this is only a necessary condition. Observing that both directions, $\Longrightarrow $ AND $\Longleftarrow $ hold makes it unnecessary to check whether your finding $x=-1/2$ is actually a solution. This may be obvious in a case like this, but it isn't if you divided by a variable and haven't ruled out that it could have been zero, or if you square an equation. E.g. $x+1=0 \Longrightarrow x^2=1 \not\Longrightarrow x=1.$ The more complicated your algebraic manipulations are, the more is it necessary to check whether something found at the end of the calculations is actually a solution to the problem. By writing $\Longleftrightarrow $ you make sure that you checked the opposite direction, too.

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  • $\begingroup$ You do not need to check the reverse implication in this case (although it happens to be true) . The equation on the left is sufficient to deduce the value of $x$. The one way implication finds the solution with no ambiguity. Of course it is always good practice to check your final answer in the original equation, but in this case there is no logical need to do so. $\endgroup$ Mar 4 at 1:13
  • $\begingroup$ @EthanBolker Sure. I just wanted to teach the principle as it is often simply ignored. To check the necessity, however, is a necessary step, even if the answer is no. Believe it or not, I once had a thesis on my table where a main argument was $ab>cb \Rightarrow a>c$ and $b<0$ was not only not investigated, it was also absolutely possible. (Of course, $a,b,c$ were algebraic expressions themselves, but the basic logic was as I have said.) $\endgroup$ Mar 4 at 15:36

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