Bob claims to have a proof that $0.\dot1=1$.
That's $0.\overline1=1$, $0.(1)=1$ or $0.11111...=1$ in other common formats.
The proof starts $$\text{If }1x=0.\dot1,\\ \text{then }10x=1.\dot1\\ 10x-1x=1.\dot1-0.\dot1\\1x=1\\ \text{substituting in the value of }1x\text{ for }0.\dot1\text{ (as defined at the start)}\\ \\0.\dot1=1$$

He is not wrong (Ignore the title). Everything is correct. Every number in this question is in base $10$.

How is this possible?


"There are 10 types of people in this world, those who understand binary and those who don't."

Bob is doing his calculations in base 2 (aka. binary): $$0.111..._2 = 1_2$$ similarly to the the following in base 10: $$0.999..._{10} = 1_{10}$$ The apparently wrong part is correct when the calculation is done in base 2: $$10_2 - 1_2 = 1_2$$ The last sentence states that every number is in base 10, which interpreted correctly (as a binary number again) means that every number is in base $$10_2 = 2_{10}$$

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  • $\begingroup$ Absolutely right! Well done. $\endgroup$ – Artyer Mar 19 '16 at 13:34
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    $\begingroup$ if that were so then every number in every base would be in base 10 (ha ha) since that's how you write the base in the base. $\endgroup$ – Kate Gregory Mar 19 '16 at 17:47
  • $\begingroup$ @KateGregory Yes, but the critical part works only for one base. $\endgroup$ – Sleafar Mar 19 '16 at 17:49
  • $\begingroup$ @KateGregory Yes. Actually I found that extra amusing and nice lateral thinking ("base ten" in numbers). $\endgroup$ – Oliphaunt - reinstate Monica Mar 19 '16 at 19:26
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    $\begingroup$ @Sleafar That's the joke: recursive base annotating. $\endgroup$ – Kroltan Mar 21 '16 at 0:40



Is wrong. It should be:


Then everything works out correctly.

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  • $\begingroup$ No, it's not wrong. Bob is doing something different. You have to read the question more deeply to find the answer. $\endgroup$ – Artyer Mar 19 '16 at 10:56

In Bob's system of math, the symbol '-' does not mean subtract; instead, it means divide, then take the base 10 log.

Therefore, 10x - 1x = (10-1) * x = log(10/1) * x = 1x (LHS) and 1.111... - .111... = log(10/9 * 9/1) = 1 (RHS)

Since step 3 to step 4 is the only "error" in the proof, the proof is now correct.

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  • $\begingroup$ Wouldn't it be (For the left) $10x-1x=\log_{10}(\frac{10x}{1x})=\log_{10}(10)=1$, and not $1x$? $10x-1x\neq (10-1)\times x$ with this meaning of subtract. $\endgroup$ – Artyer Mar 20 '16 at 9:11

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