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So, my problem is about:$\frac{(1-1)(1-1)}{(1-1)}$ I find two different ways of solving it with two different results:

-1) $$\frac{(1-1)(1-1)}{(1-1)} =\frac{(0)(0)}{0} =Undeterminate$$ In this case I first solved that what is in parenthesis.

-2) $$\frac{(1-1)(1-1)}{(1-1)} =[Step1]\frac{(1-1)^2}{(1-1)^1} =[Step2]\frac{(1-1)^{2-1}}{(1-1)^{1-1}} =[Step3]\frac{(1-1)^1}{(1-1)^0} =[Step4]\frac{(1-1)}{1} =(1-1) =0$$ In this case I waited till the end to solve that what is in parenthesis.

So I think that the second way goes wrong but my question is why? I think that it is because there must be some order betwen sovling that what we can obviously solve inside parenthesis and playing with the power of it. But, if that's the case, in which step of the calculation it started to be wrong?

-Edit: I added some step numbers $[StepX]$ to the second way to solve so that you can express well at witch step is the mistake for you and why.

-Edit: I exept an answer about where solving that what is obviously solvable in parenthesis must be solve in priority and where you can delay it. If your answer is not about parenthesis priority, please explain me first why?

-Edit: I used $(1-1)$ to express something that is equal to $0$ but you are not supposed to know that the result is $0$ before you decide to solve it on a specific step. If it is difficult for you, you can replace it by $(47895-34576-13319)$ witch is less obviously $=0$ unlsess you decide to engage in a in betwen parenthesis calculation.

lazare
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    The second way is invalid because the equation $$\frac{a \times b}{a} = b$$ is valid if and only if $~a \neq 0.~$ Division by zero is simply not allowed. – user2661923 Oct 14 '24 at 20:12
  • How do you know if 1-1=0 if you don't solve it prior? – lazare Oct 14 '24 at 20:14
  • You do solve it prior. In the original expression to be computed, $$\frac{(1-1) \times (1-1)}{1-1},$$ the first thing to do is evaluate the denominator. Once the evaluation is that the denominator equals $~0,~$ you then immediately know that the analysis in my previous comment is pertinent. – user2661923 Oct 14 '24 at 20:17
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    There are many questions on this site that discuss the trouble you encounter when trying to use formal arguments that quietly turn out to include division by $0$. So this question is essentially a duplicate of something here. (I am surprised that your last two essentially elementary questions come from someone asking also about much more advanced mathematics.) – Ethan Bolker Oct 14 '24 at 20:18
  • As an example where this issue can show up in practice, suppose $f(x)=(x-1)^2/(x-1)$. Then $f(x)=x-1$ unless $x=1$, and thus $f(x)\to 0$ as $x\to 1$. But nonetheless $f(x)$ has a (removable) singularity at $x=1$. – Semiclassical Oct 14 '24 at 20:23
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    This question is similar to: Clarification on $\frac{0^n}{0}$ when $n>0$. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Martin R Oct 14 '24 at 20:26
  • @MartinR Maybe, at least in the intentions, I think here is different since here the issue and main confusion for the OP is with the role of parenthesis. – user Oct 14 '24 at 20:34
  • I edited my post. – lazare Oct 14 '24 at 20:35
  • You need to specify how you approach to this limit: are there some variables tending to $\pm 1$? Your original equation is undefined as you are dividing by $1-1=0$ which has no multiplicative inverse: $\nexists \frac{1}{0}\in\mathbb{R}$. – Dr Potato Oct 14 '24 at 20:41
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    Dr potato, you are talking about the first way to solve I showed. Plus for everyone: I edited again: the answers must orbit around the topic of parenthesis priority. Thant you. – lazare Oct 14 '24 at 20:51
  • Parentheses or not, the expression is undefined as the denominator is zero "in disguise". Of course there are many ways to express zero for example by $(111-111)$ but that doesn't matter the "value" is still zero. It does not make sense to do algebraic manipulations on something undefined. By the way $(1-1)^{(1-1)}=0^0$ is undefined too and it is not equal to $1$ (you can find many post about it on this site). – User Oct 14 '24 at 21:15
  • Martin R your question have nothing to do with mine oviously. Please can you delete your post before the ia that manage this website close my question. If you don't like my question, why don't you just just downvote a pass your way instead of trying to close and eventualy destroy the efforts of others randomly. – lazare Oct 14 '24 at 21:19
  • You can't divide by zero. Period. – fleablood Oct 14 '24 at 22:59
  • "I waited till the end to solve that what is in parenthesis" The results not going to change no matter how long you wait. $1-1$ is $0$ whether you solve it or not. And you can not divide by zero. Ever. Period. – fleablood Oct 14 '24 at 23:01
  • BTW $\frac {x^3}x \ne x^2$ for all $x$ and $\frac {(x^2 -3x -10)^7}{(x^2-3x-10)^5}\ne (x^2-3x-10)^2$ for all $x$ either. If $x=0$ then $\frac {x^3}{x}$ doesn't equal anything and if $x=5$ or $x=-2$ then $\frac {(x^2-3x-10)^7}{(x^2-3x-10)^5}$ doesn't equal anything either. You can't just put a mask over your head and say "If I don't know this is zero, I can pretend this is okay". If you get a problem like $x^3y^5 + x^2y = 9xy$ you can NOT (I will not let you!) divide both sides by $xy$. You can't do that BECAUSE you don't know whether $xy=0$ or not. – fleablood Oct 14 '24 at 23:11

1 Answers1

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The issue is with step

$$\ldots =\frac{(1-1)^2}{(1-1)^1} =\frac{(1-1)^{2-1}}{(1-1)^{1-1}}=\ldots$$

since we are dividing by $(1-1)=0$ which is not allowed.

Moreover you are also using that $0^0=1$ which is also wrong (at least in this context).

Refer also to:

With the second solution we have as a condition $(1-1)\neq 0$ and since at the end we end up with $(1-1)=0$ as solution, all the derivation is invalid and we need to discard that solution.

To be more clear, for the second approach, let indicate $(1-1)=x$, we should proceed as follows, the expression

$$\frac{x\cdot x}x$$

is well defined only when $x\neq 0$, in this case we have

$$\frac{x\cdot x}x=\frac{x\cdot \frac x x}{\frac x x}=\frac{x\cdot 1}1=x$$

but $x=(1-1)=0$ therefore the solution is not acceptable.

user
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  • At this step we don't know that what is the result of (1-1) because we decided to solve it later, it could be $5$ or any , who know? If you delay to solve it .. – lazare Oct 14 '24 at 20:18
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    @lazare You can delay sure but we must assume as a condition $(1-1)\neq 0$ which is false. – user Oct 14 '24 at 20:19
  • @lazare Essentially the same comment that you left for me, following your posted question. See my responding comment. – user2661923 Oct 14 '24 at 20:20
  • In other words, similar to what already explained by user2661923, you are computing $\frac{x\cdot x}{x}=\frac{x\cdot 1}{1}=x$ which is fine only for $x\neq 0$. – user Oct 14 '24 at 20:24
  • I edited my post. – lazare Oct 14 '24 at 20:35
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    So you say $[Step2]$ but it sound like you talk about $[Step1]$ because in $[Step1]$ there is already the problem of verifying if the denominator is equal to zero or not. Plus I edited again. I exept that the answers stay on the topic of parenthesis priority. – lazare Oct 14 '24 at 21:01
  • @lazare Yes of course from the start we need that condition since $\frac{x\cdot x}{x}$ requires $x\neq 0$, in this case if you have already stated this condition Step2 if fine but things don't change since at the end we come up with $x=0$ which is not acceptable. – user Oct 14 '24 at 21:05
  • I don't understand why you are still talking about $[Step2]$ for a problem found in $[Step1]$ ... ? I understand clearly that the result is not acceptable but the main question is not about that – lazare Oct 14 '24 at 21:11
  • @lazare I've eliminate indeed any reference to Step2, I've added another explanation. The issue is not with parenthesis, you are allowed to do all the steps but at the end we need to check that $(1-1)\neq 0$ which is false and so the derivation is invalid. – user Oct 14 '24 at 21:16
  • So if understand that what you added: the in betwen parenthesis of (1-1) must be solve at $[Step1]$ not becauce of parenthesis priority over power changing but beceause we have to check if the denominator is not equal to zero before going further? (And why you don't use the step indication?) – lazare Oct 14 '24 at 21:27
  • Yes the issue is not with the parenthesis priority, if we don't know the value for $x=(1-1)$ all steps are fine . – user Oct 14 '24 at 21:40
  • @lazare Maybe thigs are more clear for an equation such $(x^2-1)/(x-1)=0$ for which we need $x-1\neq 0$ as initial condition, then we have $$(x^2-2x+1)/(x-1)=0\iff ((x-1)/(x-1))/(x-1)=0 \iff x-1=0$$ which is not acceptable. – user Oct 14 '24 at 21:43
  • So the $[Step1]$ must be searching for initial conditions before going further? – lazare Oct 14 '24 at 21:51
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    @lazare For any expression as first step we need to set the condition in order to have a well defined expression, as for example for $\sqrt x $, $\log x$, $\tan x$, and so on. – user Oct 14 '24 at 21:53
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    I would say that this is a Step0. – user Oct 14 '24 at 21:53
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    @lazare You don't have to search for initial conditions, but if you do you must note what would cause the problem for the rest of the reasoning. It's fine to cancel an $(x-1)$ from the denominator, but then you must note that all the remaining steps are only true if $x-1 \neq 0$. Similarly, you can cancel a $(1-1)$ from the denominator as you do in step 3, but then you must have a note that the rest of your steps are only valid if $1-1 \neq 0$. It's fine to wait until the end to check what these conditions mean, though in the case of $1-1$ you might save some effort to do so earlier. – MartianInvader Oct 14 '24 at 22:12
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    Well, first of all.... Even saying $\frac {x^2}{x}$ isn't allowed unless you specify you are only talking about cases where $x \ne 0$. Saying something like $\frac {x^2 -3x+2}{x-1}= 7$ is meaningless unless we clarify that $x-1$ can not be equal to $0$. If a problem says "$\frac {x^2-3x+2}{x-1}=7$" then we know $x-1\ne 0$ because otherwise $\frac {x^2-3x+2}{x-1}$ is meaningless garbage and can't equal $7$ as it is doesn't equal anything. – fleablood Oct 14 '24 at 23:39