Digit Manipulation part 2

Still… between this blog and the YouTube video, there doesn’t seem to be a good explanation for why the answer to a digit manipulation comes to the magic “1” proposed by other proofs instead of some other number.

To recap:

x = .999…

10x = 9.999…

9x = 9

x = 1

The answer I’ve given for this is that the .999… after the 9 in the second step is not as long as x actually is, and subtracting x from both sides should actually read 9x = 8.999…91 instead. But why, when we do the math incorrectly, does it give us 1? Consider this: what proponents of the theory that .999… = 1 suggest is that when you multiply .999… by 10 you should get the same result that you’d get from adding 9 (9 + .999… equals 9.999… agreed?). Is it possible to get the same result by both adding 9 and by multiplying 10 to .999…? What is x given that 10x = x + 9?

10x = x + 9

9x = 9   (subtract x from both sides)

x = 1   (divide both sides by 9)

“Wow!” you exclaim. “See? x does equal both 1 and .999 because you can do the same math to both and get the same results!” If you really think this, then you have confused cause and effect — by multiplying one side of the equation by 10 and by adding 9 to the other side, you’ve actually then changed the answer to 1. To prove this, let’s change x to 2 in a digit manipulation by the same method, which is by adding 2 to one side while multiplying the other side by 2 (because given 2x = x + 2, x = 2).

x = .999…

2x = 2.999…  (after multiplying by 2 on the left and adding 2 on the right)

x = 2  (after subtracting x from both sides)

See how we made x equal what we wanted it to equal? Let’s try to make x = 3 by multiplying one side by 2 and adding 3 to the other side (because given 2x = x + 3, x = 3).

x = .999…

2x = 3.999…  (after multiplying by 2 on the left and adding 3 on the right)

x = 3  (after subtracting x on both sides)

Do you see the pattern here? This is why x = 1 in the original equation… because we actually added 9 to .999… instead of multiplying by 10. It gave us an end result that we can predict. Let’s try this one last time, except instead of .999… we’ll use .777… to show that this isn’t a trick that is only true with .999…

x = .777

10x = 7.777…

9x = 7

x = 7/9 which almost equals .777…  (see earlier post on why it “almost” equals that).

 

Given 10x = x + 7, what is x?

9x = 7  (after subtracting x from both sides)

x = 7/9  (after dividing both sides by 9)

 

See? It’s like a self-fulfilling prophecy.

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About starcrashx

I love statistics. They drive my poker playing, my reasoning, and my research. As Penn Gillete said "Luck is probability taken personally". There's no such thing as luck... but I wish you positive chance.
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3 Responses to Digit Manipulation part 2

  1. Wani says:

    So, in short, your argument here is “if I do maths wrong, I can get any answer I like! Therefore, maths is wrong!”

    You cannot multiply by 2 on one side of an equation, then add 2 on the other side. Your equations are then no longer equal, and you need to change your sign to in inequality sign.

    The first equation (x = 0.999…, multiply by 10 to get x = 9.999…) is not just adding 9 to the left, it’s quite clearly a valid multiplication by 10.

    • starcrashx says:

      I’m well aware of the fact that I “cannot multiply by 2 on one side of an equation, then add 2 on the other side”. I think you missed my point. I was demonstrating why the answer to the digit manipulation is more consistent with adding 9 to the right side of the equation than multiplying by 10.

      All the same, you didn’t entirely miss my point. Yes, by doing the maths wrong, you can get any answer you like. And the popular way of doing the first equation is wrong, which is why it produces the wrong answer.

  2. Matthias says:

    What is .999…? Just give me some definition!

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