The_root_of_all_evil said:
Maze1125 said:
I gave you an entire thread full of proofs.
Which I disagree with because you're using your own stated proof as part of the solving equation. That's paradoxical in itself.
What? That doesn't make sense in context,
at all.
I gave you a thread full of proofs, all of which were correct independently.
So how on Earth does that lead you to conclude that "you're using your own stated proof as part of the solving equation."
But hey, seen as you didn't bother to read my post last time.
See, that's what the problem is. I actually read it - found the problem - and discarded it. You're the one who is being aggressive about YOUR proof, because the onus of proof relies on you.
If you found the problem, why didn't you mention it?
And, as I said, the thread was
full of proofs. I'd like to see your problem with each one of then.
Now sum(from k=1 to n) (9 * 1/10^k) is a finite sum - No proof available.
What? How you do even require a proof of that. a 4 year old could see that was finite.
Anyway, I'll do it for you anyway.
n is finite. There are n terms in the sum, therefore there are a finite number of terms in the sum.
For each k > 0, 0 < 9 * 1/10^k < 1, therefore each term is finite.
Every term is finite and there are only finitely many terms, therefore the sum is finite.
, and so we can calculate that
|1 - sum(from k=1 to n) (9 * 1/10^k)| = |1/10^n| - Assumption - Working not shown.
Okay, they
literally teach you how to add and subtract decimals when you're 12. And you expect me to believe that you've done 4 years at university when you can't even subtract a decimal from 1?
Oh well, I'll try and teach you what your school obviously failed to.
For k=1
) 1.0
)-0.9
(I'm using the ")" to maintain the format)
First we subtract the right most terms.
Unfortunately 9 > 0, so what we do is "borrow" a 1 from the next column. Allowing us to do 10 - 9 giving us a 1 in that column. And it turns out that we borrowed everything that was available in the left most column, leaving us with 0 - 0. So, the final result is:
) 1.0
)-0.9
)-------
) 0.1
For k=2 we have:
) 1.00
)-0.99
This time with have to borrow across two columns, which still gives us the 10 - 9 in the right most column, and leaves us with 9 - 9 in the middle column and 0 - 0 in the left most, so the result is:
) 1.00
)-0.99
)--------
) 0.01
Now, for k = n, we have to borrow across n columns, but the principle still works, Giving:
) 1.00...00
)-0.99...99
)-------------
) 0.00...01
(Note in this case the ... stop with another number, indicating the chain is finite.)
Now, if you have any more problems, please take them up with your 7th grade maths teacher.
Try proving that e=0 from first principles.
What? You never need to prove e = 0. In fact, e cannot be 0, it is specifically designed to be any positive number that
isn't 0.
I have to wonder is you've ever done a limit proof before.
And as for first principles, I proved it using nothing but finite addition and subtraction, limits and the definitions. How much more first principle could I get?