How does it show why 1.9 != 2? You can show it algebraically.
1.9 = 1 + 0.9
Let 0.9 = X, then
10X = 9.9
9X = 9.9 - 0.9 = 9
X = 1
We know 1.9 = 1 + 0.9, so
1.9 = 1 + X = 1 + 1 = 2
How does it show why 1.9 != 2? You can show it algebraically.
1.9 = 1 + 0.9
Let 0.9 = X, then
10X = 9.9
9X = 9.9 - 0.9 = 9
X = 1
We know 1.9 = 1 + 0.9, so
1.9 = 1 + X = 1 + 1 = 2
As stated before, an easy way to visualize 0.9 = 1 is
1/3 + 1/3 + 1/3 = 1
1/3 = 0.3
0.3 + 0.3 + 0.3 = 0.9
Thus 1 = 0.9
So if we do the same for 2:
2/3 + 2/3 + 2/3 = 2
2/3 = 0.6
0.6 + 0.6+ 0.6 = 1.8
1.8 = 2, but you can see that 1.8 isn't anywhere close to two.
Then again this is why the above 1/3's and 0.3's proof is a simplistic way of looking at 0.9 = 1, as it's not an accepted proof, just an easy way to visualize it. The same is true for 1.8 != 2. It's proven in a geometric proof, and the above is a simplistic way of looking at it, even if it isn't 100% concrete.
That's what algebraic proofs are for
1.8 != 2 however, no matter how you look at it. It's a result of adding the 0.6's incorrectly. 1.8 is 1 + 8/9, which you can clearly see is not 2, so you must have assumed or done something wrong in your previous addition.
Oh, guess I misunderstood you then![]()
Why can't you do it though? They might be infinite, but they're also all defined and known and known to be the same at every single place value, so multiplying by whatever base the numbers are in at the very least (so multiples of 10 in this case) would be safe, since all you'll be doing then is shifting the location of the decimal point. I'm not sure if multiplying by other numbers would be safe though, not without more careful thought being put into the process to make sure you don't introduce errors by mistake.
Also, you can't assume that 0.9 = 1 in this proof. I don't remember the exact reason (maybe someone here does), but proving that 1.9 = 2 is like proving that 0.9 = 1 so you can't use the 0.9 = 1 assumption.
EDIT: Heh, as I was searching the web for 1.9 =/!= 2 info, I found the proof that you are using above.![]()
I didn't assume 0.9 = 1 though, I proved it first before sticking it in the last equation
If you want though, we can do it another, but similar, way.
Let X = 1.9
10X = 19.9
9X = 19.9 - 1.9
9X = 18
X = 18/9 = 2
Okay, but here's a counter argument/proof:Originally Posted by Arche
1.9 = 1 + 0.9
Let 0.9 = X, then
10X = 9.9
9X = 9.9 - 1 = 8.9
X = 8.9/9 = ??
Either way once you hit this point dividing into an infinite number is not easy to do.
See, in your proof you are inter-changing 0.9 and 1. All if takes for me to prove this proof wrong is to interchange the 0.9 and 1 in a couple of different places. If your proof were true then it would have the same answer no matter where 0.9's or 1's are used.
Then I'd just go on to show you that 8.9 = 9. You haven't actually given a counter example
EDIT: Also there's a flaw in your counter-argument. How can you subtract 1 from 10X to get 9X when you haven't proven X = 1 yet? I can interchange them freely in mine though, but only after I've proven 0.9 = X = 1 already.
Er, I subtracted X from the LHS, which I said was 1.9 in an earlier line, so there's nothing wrong thereOriginally Posted by Flying Mullet
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I did subtract 0.9 from the right thoughOriginally Posted by Arche
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