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I was reading a book on mathematical logic, and I was confronted with the following question:

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$\sqrt{2x+1}=\sqrt x-5 ⇒ 2x+1=(\sqrt x-5)^2$

$⇔ 2x+1=x-10\sqrt x+25$

$ ⇔ 10\sqrt x=24-x$ $ ⇒ 100x=(24-x)^2$

$ ⇔ 100x=576-48x+x^2$

$ ⇔ x^2-148x+576=0$

$ ⇔ (x-4)(x-144)=0$

$ ⇔ x=4 ∨ x=144$

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Since, ⇔ means logically equivalent, I cleared up the question and I reached:

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${\sqrt{2x+1}=\sqrt x-5 \quad ⇒ \quad 10\sqrt x=24-x \quad ⇒ \quad x=4 ∨ x=144} \quad → $

$\sqrt{2x+1}=\sqrt x-5 \quad ⇒ \quad (x=4 ∨ x=144)$

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At this point the book said "since the solution doesn't agree with the equation, then it must be false; and therefore, so is the equation itself"

My main problem with this explanation was that I thought:

The equation is implying a certain solution; however, the solution doesn't imply the equation (hence the use of the forward implication sign). This means that the two statements are not logically equivalent.

So, how can we decide whether the statements are true or false? What is our reference to deciding whether the solution is true (do we just say, that the solution is a false statement just because it doesn't agree with the equation whose truth value itself is unknown)?

Can't an equation such as the one above be of the case, false equation $ ⇒ $ true solution (I actually couldn't make sense out of the last situation I suggested, which I believe is technically a possibility; my conclusion was that this can only happen when the solution is actually a true statement which is totally unrelated to the question and in deduced not from the equation)?

I'd be grateful if you could help me with this problem of mine.

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  • $\begingroup$ If the equation had a solution, it would be $4$ or $144$. But those aren't valid solutions, so the assumption that the equation has solutions is false. See also contraposition. $\endgroup$ – Jaap Scherphuis Aug 7 '18 at 12:36
  • $\begingroup$ If $p $ implies $q $ ($p\rightarrow q $ is true ) and $q $ is true, then $p $ could still be either true or false. Eg $p:-3x=3x $, $q:9x^2=9x^2$. $\endgroup$ – AnyAD Aug 7 '18 at 12:57
  • $\begingroup$ See this link. Also, note: $\sqrt{x}\ge 0 \Rightarrow x\ge 0 \Rightarrow \sqrt{2x+1}+5>\sqrt{x}$. $\endgroup$ – farruhota Aug 7 '18 at 14:00
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    $\begingroup$ Could you please give the details (e.g. author and title) of the logic book you are using? Not many get into application of logic to this sort of everyday textbook math. thanks. $\endgroup$ – Cris P Aug 7 '18 at 19:21
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The given equation defines a solution set $$S:=\bigl\{x\in{\mathbb R}\bigm|\sqrt{2x+1}=\sqrt{ x}-5\bigr\}\ .$$ Then followed a chain of propositions connected with $\Rightarrow$ and $\Leftrightarrow$, as follows: $$x\in S\Longrightarrow\ldots\Leftrightarrow\ldots\Longrightarrow\ldots \Leftrightarrow\ldots\Leftrightarrow \bigl(x=4\vee x=144\bigr)\ .$$ This proves $S\subset\{4, 144\}$, no more, no less. In order to determine $S$ completely we now have to check which of $4$ and $144$ satisfies the original equation. It turns out that none of them does. Therefore we can safely say that $S=\emptyset$.

There is nothing false here. The original problem posed an innocently looking condition, but this condition cannot be fulfilled within ${\mathbb R}$; that's all.

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If you try to solve an equation of the type$$f(x)=g(x)\tag1$$by squaring both sides, then what you are acutally doing is to solve the equation$$\bigl(f(x)\bigr)^2=\bigl(g(x)\bigr)^2.\tag2$$But the solutions of $(2)$ don't have to be solutions of $(1)$. If $x$ is such that $(1)$ holds, all you can say is that $f(x)=\pm g(x)$. So, although it is correct that every solution of $(1)$ must be a solution of $(2)$, it is not always true that every solution of $(2)$ is a solution of $(1)$. Whether or not this occurs must be checked case-by-case.

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