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I have seen such proofs many times and was unable to prove where it was wrong (not a math person-my bad).
For example the following proof for $\frac{0}{0} = 2$ looks like it is proven correctly - but somehow I feel that it cannot be right.

enter image description here.

This could be pun but I want to know where it is wrong

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    $\begingroup$ It is wrong when in the beginning ou assume that $\frac 00$ is defined. $\endgroup$ Jun 6, 2016 at 18:10
  • $\begingroup$ It involves canceling a $(10-10)$ in middle, which is forbidden, as $(10-10)=0$. $\endgroup$ Jun 6, 2016 at 18:11
  • $\begingroup$ It is especially wrong when you divide the top and bottom of the fraction by $0$ $\endgroup$ Jun 6, 2016 at 18:11
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    $\begingroup$ Any argument that uses the expression $\frac{0}{0}$ is wrong. Sometimes a wrong argument happens to yield a correct answer. $\endgroup$ Jun 6, 2016 at 18:11
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    $\begingroup$ $0 \cdot 4 = 0 \cdot 1 \Rightarrow 4 =1 $ by cancelling out the zeros. Which is clearly nonsense. The laws of maths specifically disallow dividing by zero. So no this is not a proof. There are lots of false proofs which rely on hiding a division by zero. $\endgroup$ Jun 6, 2016 at 18:18

2 Answers 2

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The error starts right at the beginning, it is always wrong to divide by zero. Any conclusion can be followed from wrong assumptions.

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What is wrong is that you can't consider $\frac 00$ in the first place. So everything you will do after that will just be false, because it's from something which is false.

You can show the same way that $\frac 00=3$, and this is nonsense.

Plus, you can't not simplify by $10-10$, because you can never simplify by $0$.

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    $\begingroup$ If you start from something meaningless, you might infer other things that are meaningless, or you might infer some things are are meaningful. The meaningful statements might be true, but they could be false. $\endgroup$
    – hardmath
    Jun 6, 2016 at 18:14
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    $\begingroup$ I couldn't agree more ! $\endgroup$
    – E. Joseph
    Jun 6, 2016 at 18:18

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