Math Is Fun Forum

The site works here, and I think I've seen it before too, heh.

Have you seen the record list, Devanté? I know there's a list with people who holds some kind of record in memorizing pi digits somewhere online. I don't know if it was official, and I don't have the site url, but try searching google and you might find it. It was pretty insane, but I'm pretty sure it wasn't close to a million though!

I've had a calculator that gave me the wrong answer once (for real! and I threw it away too!), so I don't always trust them
Learn the trick John described, sooner or later you'll be using it for polynomial division anyway

If that proof is true, then 3,999... is in Q.

If 1,999... = 2, then we could multiply both by 10000...
thus, 1999... = 2000... right?

Actually, I don't think I'll continue from here, I'm really confused now...

Yeah, true that, and the prefixes are important. Here's how I'd do it:

Split up the same way Ricky did and introduce a variable A for simplicity

Now we have:

From here, you can simplify the root (check it yourself):

Is (...) supposed to mean limit? Now that's new.

Besides, I think this would be a valid argument against it:

Edit: 3,999... is not irrational by definition so it has to be rational, so I take the above statement back.

The point is that if you prove that if it is true for n=k, it is also true for n=k+1. If we end up with a statement which is true based on that assumption, it means that the assumption we made is true. If, however, we'd end up with a false statement as a result of that, our original assumption would be wrong.

You can look at it this way:

1) If x is true --> y is also true
2) If x is true --> y is false

1) If our assumption x is true, it leads to a statement y which is true. So our assumption must be correct.
2) If our assumption x is true, it leads to a statement y which is false. So our assumption cannot be correct.

Apply this to your problem. If what you are saying is true, two of those formulas should lead to a false statement, meaning that the assumption that it is a formula for the sum of natural numbers is false.

solved the first 4 :] actually, the first two were just naked singles all through, no fun

Nice John, 12^(3 1/3) is a much better answer :]

Hmm, ok I'll take it as my first suggestion then and lets see what I get.

We have that

The second equation says that L has increased 1% (1,01) and gamma is the change in K, which we should solve. Set them equal to get rid of Q:

Since

So,

Which is approximately 0,99504... in other words a decrease by 0,5%.

1. x=25/10=5/2. [divide by ten on both sides]

2. 10e^x=160 -> e^x=16 -> x=ln(16) [subtract 40, divide ten, logaritm both sides and ln(e)=1]

3. ln(x^0,3)=ln(12) -> ln(x)=ln(12)/0,3 -> x=e^(ln(12)/0,3)  [logaritm both, down with 0,3; solve for x]

4. ln(x)*ln(e)=ln(8) so ln(x)=ln(8), thus x=8. [logaritm on both sides, ln(e)=1]

5. Do you mean Q=K^(0,8)*L^(0,4) or Q=K^(0,8(L)^(0,4)) or something else?

Welcome Ross!

Your swedish is fine, lol. I hope you enjoy your stay here :]

Being swede, I'm really looking forward to the game against england today