I want to understand if I understand the concept of equation correctly. For this I want to know if $0$ is the root of the equation $\frac{x^2}{x} = 0$. If we simplify the equation then we can get equation $x = 0$ and then we have solution $0$. But if we replace $x$ by $0$ in the original equation then we get expression $\frac{0^2}{0} = 0$ which doesn't make sense. I'm interested in your opinion.
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5The equation has no solution since the left side is $x$ for $x \ne 0$ , but undefined for $x=0$ – Peter Jun 24 '22 at 11:23
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Thank you @Peter – user341 Jun 24 '22 at 11:24
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2See posts like this one concerning division by zero. – Dietrich Burde Jun 24 '22 at 11:26
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Also, an equation can have solutions, a function roots. An equation has no root. – Peter Jun 24 '22 at 11:28
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The equations $x^2/x=a$ and $x=a$ have different definitions. The former is not defined when $x=0$ – NadAlaba Jun 24 '22 at 11:36
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It is worth pointing out that the function $f(x) = \begin{cases}\frac{x^2}{x}&x\neq 0\42&x=0\end{cases}$ (I just picked 42 as an example, no real mathematical meaning to the number 42... I just needed the domain to be all real numbers) has a removable discontinuity at $x=0$. That is likely what you are intuitively trying to get at in attempting to simplify the fraction and cancel an $x$ from top and bottom. The punchline is that officially the original function you refer to is not defined for $x=0$, that's outside the domain – JMoravitz Jun 24 '22 at 12:16
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but your function does have the limit as $x\to 0$ of zero. – JMoravitz Jun 24 '22 at 12:17
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@JMoravitz 42 actually seems to have a meaning… https://en.wikipedia.org/wiki/Phrases_from_The_Hitchhiker%27s_Guide_to_the_Galaxy#The_Answer_to_the_Ultimate_Question_of_Life,_the_Universe,_and_Everything_is_42 – insipidintegrator Jun 24 '22 at 14:48
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@insipidintegrator yes, that was the inside joke... it didn't need to be said aloud. We just commonly use $42$ when we have a choice available to make and the number used doesn't actually matter to prove the point we were trying to make... – JMoravitz Jun 24 '22 at 14:51
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No, this system is indeterminate and there is no solution for $x=0$. The key point in this question is to figure out how simplifying a system can often hide certain properties of it, or how simplifications can be incorrectly made by violating certain mathematical conditions.
Take for instance, the equation you have: $$\frac{x^2}{x} = 0$$
All mathematical systems are dependent on how they are simplified. In this form, there can be no solution, because have the form $\frac{0^2}{0}$ which is indefinite.
The problem with the method of simply bringing the denominator over to the right-hand side (RHS) is that it simplifies the equation and loses the importance of the fraction representation. The indefinite form specifically has a condition that if it is of the form $\frac{0}{0}$, then it cannot be simplified by shifting the denominator to the opposite side (LHS or RHS). Thus, when simplifying the system, you missed this important condition.
Bottom line, canceling 2 terms on the RHS and LHS require the term to be divided across both sides, but this yields an indefinite answer if the division creates an indeterminate/indefinite form.
Here's a famous example of a proof that goes wrong because of the incorrect application of the above condition: $$ \begin{align} Let, \ a & = b \\ a^2 & = ab \\ a^2 + a^2 & = a^2 + ab \\ 2a^2 & = a^2 + ab \\ 2a^2 - 2ab & = a^2 + ab - 2ab \\ 2(a^2 - ab) & = a^2 - ab \\ 2 & = 1 \end{align} $$
The error lies in the last step, since you cannot cancel out the term $a^2-ab$ from LHS & the RHS. The reason is because when you cancel out the term, you are essentially dividing the term across both sides. However, you know that $a^2 - ab = 0$ from the assumption $a=b$, meaning that your equation becomes: $$ \begin{align} 2\left[\frac{a^2-ab}{a^2-ab}\right] & = \left[\frac{a^2-ab}{a^2-ab}\right] \\ 2\left[\frac{0}{0}\right] & = \left[\frac{0}{0}\right] \end{align} $$ Which is an indeterminate/indefinite form, rendering the proof incorrect. The same logic extends to your question.
Ajay Menon
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No, the equation has no solutions. The problem comes from the step of simplifying the equation. In this step you are dividing by $x$, but in that case you are dividing by $0$, which is not allowed. If you assume that $x\not=0$, you can simplify and get $x=0$, which is a contradiction. So the equation has no solutions
