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As we know that $$\frac{1}{0} = ∞ \implies \frac{1}{∞} = 0 \implies 1 = 0.∞$$ Now let $x=0$ and $y=0$ then $$x+y=0 \implies \frac{x+y}{1}=0 \implies \frac{x+y}{0}=1$$ and $x+y=0$ so $$\frac{0}{0}=1$$ Similarly, as $\frac{2}{2}=1$ and $\frac{5}{5}=1$

Now we know that any number multiply by zero is equal to zero so $$ (2)0 = 0 \rightarrow(1)$$ And $$\frac{0}{0} = \frac{0}{0}$$ By eq(1) $$\frac{(2)0}{0} = \frac{(1)0}{0}$$ $$2\frac{0}{0} = 1\frac{0}{0}$$ as $\frac{0}{0} = 1$ then $$2(1) = 1(1)$$ $$2 = 1$$

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    $\begingroup$ Division by zero is not defined. It could be anything from $-\infty$ to $+\infty$ as well as any complex number. $\endgroup$ – Ruslan Feb 13 '15 at 10:26
  • $\begingroup$ math.stackexchange.com/questions/417280/… $\endgroup$ – Alex Feb 13 '15 at 10:26
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    $\begingroup$ I suggest that this question be placed under the "recreational mathematics" category. $\endgroup$ – Vim Feb 13 '15 at 10:29
  • $\begingroup$ Useful to read: en.wikipedia.org/wiki/Division_by_zero $\endgroup$ – Ruslan Feb 13 '15 at 10:33
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    $\begingroup$ This is why analysis exists. $\endgroup$ – Arturo don Juan Feb 14 '15 at 5:07
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$\infty$ is not a number and $\frac{0}{0}$ is not determined.

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    $\begingroup$ I would add "real" in front of number. $\endgroup$ – Regret Feb 13 '15 at 10:32
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Your starting point is not at all an identity between numbers, but only a mnemonic trick for the statement:

$$ \lim_{x\rightarrow a}{f(x)}=0 \Rightarrow \lim_{x\rightarrow a}{\dfrac{1}{f(x)}}=\infty $$ Properly using the definition of limit you can show that: $$ \lim_{x\rightarrow a}{f(x)}=\infty \Rightarrow \lim_{x\rightarrow a}{\dfrac{1}{f(x)}}=0 $$ That we can ''remember'' quikly as $$ \dfrac{1}{\infty}=0 $$ But we cannot proof that $$ \left(\lim_{x\rightarrow a}{f(x)}=0 \quad \land \quad \lim_{x\rightarrow a}{g(x)}=\infty \right)\Rightarrow \lim_{x\rightarrow a}{f(x)g(x)}=1 $$ as in your $ 1=0\cdot \infty$. So all your deduction is based on the wrong idea that $ \infty$ can be treated as a number ,but it is only a symbol used to represent a particular limit.

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Here is a simpler version of your argument:

$$2 = 2 \cdot 1 = 2 \cdot \frac{0}{0} \overset{*}{=} \frac{2 \cdot 0}{0} = \frac{0}{0} = 1$$

The problematic step is highlighted with an asterisk. Every other step is okay, provided that we define $0/0$ to equal $1$. What this teaches us is that we cannot both have our cake, and eat it too; if we define $0/0$ to equal $1$, then the familiar law:

$$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$$

no longer holds.

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