Problem About Equality: Is 2=1?

Question

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 $$2(1) = 1(1)$$ $$2 = 1$$

Division by zero is not defined. It could be anything from $-\infty$ to $+\infty$ as well as any complex number. — Ruslan, Feb 13 '15 at 10:26
http://math.stackexchange.com/questions/417280/continued-fraction-fallacy-1-2/ — Alex, Feb 13 '15 at 10:26
I suggest that this question be placed under the "recreational mathematics" category. — Vim, Feb 13 '15 at 10:29
Useful to read: http://en.wikipedia.org/wiki/Division_by_zero — Ruslan, Feb 13 '15 at 10:33
Now there is another problem if x=y ⇒ x+x-x = y+y-y ⇒ 2x-x=2y-y ⇒ 2x-2y=x-y ⇒ 2(x-y)=(x-y) and by cancellation law ⇒ 2=1 — Syed Muhammad Asad, Feb 13 '15 at 13:36
This question did not deserve all those downvotes. Its only flaw is that it doesn't 100% clearly state the question (where is my mistake?). — goblin GONE, Feb 14 '15 at 16:58

score 4 · Answer 1 · answered Feb 13 '15 at 10:31

4

$\infty$ is not a number and $\frac{0}{0}$ is not determined.

answered Feb 13 '15 at 10:31

2

I would add "real" in front of number. – Regret Feb 13 '15 at 10:32

goblin GONE · Answer 2 · 2015-02-14T05:02:18.887

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.

score 2 · Answer 3 · answered Feb 13 '15 at 10:59

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.

Problem About Equality: Is 2=1?

3 Answers3