The clearly wrong proof

Question

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?

10 minus 1 equals 1????????? – Scratch---Cat Jul 25 '20 at 06:11 — Scratch---Cat, Jul 25 '20 at 06:11

Sleafar · Accepted Answer · 2016-03-19T14:04:25.537

37

"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 $$10_2 = 2_{10}$$

edited Mar 19 '16 at 14:04

answered Mar 19 '16 at 12:16

Sleafar

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Absolutely right! Well done. – Artyer Mar 19 '16 at 13:34
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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. – Kate Gregory Mar 19 '16 at 17:47
@KateGregory Yes, but the critical part works only for one base. – Sleafar Mar 19 '16 at 17:49
@KateGregory Yes. Actually I found that extra amusing and nice lateral thinking ("base ten" in numbers). – Oliphaunt Mar 19 '16 at 19:26
Every base is base 10. (And the jokes on that link are a lot of fun :) – Wildcard Mar 20 '16 at 08:39
What is the base of the base annotation? – Kroltan Mar 20 '16 at 19:05
@Kroltan 10 of course. ;) – Sleafar Mar 20 '16 at 19:13
@Sleafar 16? That's what I thought. – Kroltan Mar 20 '16 at 19:39
@Kroltan Nope, they are plain and simple decimal numbers. Otherwise I would need base annotations for base annotations ... – Sleafar Mar 20 '16 at 19:44
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@Sleafar That's the joke: recursive base annotating. – Kroltan Mar 21 '16 at 00:40
Damn! I thought the answer was beyond the binary base. I took the usage of binary base as starting point... and perhaps answer why periodic 0.111... = 1. – Luis Masuelli Jun 02 '16 at 16:30
Base 9+1 is surprisingly unambiguous however, (i think). – Samie Bee Apr 09 '19 at 05:10

score 8 · Answer 2 · edited Mar 19 '16 at 15:14

8

Well:

$10x−1x=1.\dot1−0.\dot1$
$1x=1$

Is wrong. It should be:

$10x−1x=1.\dot1−0.\dot1$
$9x=1$

Then everything works out correctly.

edited Mar 19 '16 at 15:14

Artyer

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answered Mar 19 '16 at 10:33

Paul Evans

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

score 4 · Answer 3 · answered Mar 19 '16 at 13:54

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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.

answered Mar 19 '16 at 13:54

harpematt

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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. – Artyer Mar 20 '16 at 09:11

The clearly wrong proof

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