The clearly wrong proof

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????????? Commented Jul 25, 2020 at 6:11

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

• Absolutely right! Well done. Commented Mar 19, 2016 at 13:34
• 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. Commented Mar 19, 2016 at 17:47
• @KateGregory Yes, but the critical part works only for one base. Commented Mar 19, 2016 at 17:49
• @KateGregory Yes. Actually I found that extra amusing and nice lateral thinking ("base ten" in numbers). Commented Mar 19, 2016 at 19:26
• @Sleafar That's the joke: recursive base annotating. Commented Mar 21, 2016 at 0:40

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.

• No, it's not wrong. Bob is doing something different. You have to read the question more deeply to find the answer. Commented Mar 19, 2016 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.

• 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. Commented Mar 20, 2016 at 9:11