### Re: Math proof

Posted:

**Fri May 02, 2008 2:56 pm**He said he was joking, but I don't think he was...

Training the hacker underground

https://www.hackthissite.org/forums/

https://www.hackthissite.org/forums/viewtopic.php?f=68&t=228

Page **3** of **4**

Posted: **Fri May 02, 2008 2:56 pm**

He said he was joking, but I don't think he was...

Posted: **Fri May 02, 2008 3:18 pm**

omg i was joking.......don't you read the whole post?!?!?!?!?

Posted: **Fri May 02, 2008 8:28 pm**

Ok so i know this shows how much information I don't know, but what is that series? is it a crypto algorithm or something? I google'd it and didn't find anything.

Posted: **Sat May 03, 2008 5:17 am**

krauser2288 wrote:omg i was joking.......don't you read the whole post?!?!?!?!?

I read the whole post.. I can explain why I am going to report you.

The series you posted is logic mission 28 which you haven't completed yet. And now you posted it here so that someone "by accident" can help you with it.

Posted: **Sat May 03, 2008 5:42 am**

I've decided to answer all the challenges given in this forum.

Proof for DeMorgan's law [I'll take the more general version] (A union B union C ...)^c = A^c intersection B^c intersection C^c ...

Take x in (A union B union C ...)^c.

Then x is not in (A union B union C ...).

This implies that x is not in A, not in B, not in C, ...

Therefore x is in A^c, in B^c, in C^c, ...

=> x is in A^c intersection B^c intersection C^c ...

It's a known fact that 0.99999999999...=1 (and that 0.122999999999...=0.123 and so on).

About comperr's challenge written in wikipedia code. It isn't allowed to take \infty - \infty.

Proof for DeMorgan's law [I'll take the more general version] (A union B union C ...)^c = A^c intersection B^c intersection C^c ...

Take x in (A union B union C ...)^c.

Then x is not in (A union B union C ...).

This implies that x is not in A, not in B, not in C, ...

Therefore x is in A^c, in B^c, in C^c, ...

=> x is in A^c intersection B^c intersection C^c ...

It's a known fact that 0.99999999999...=1 (and that 0.122999999999...=0.123 and so on).

About comperr's challenge written in wikipedia code. It isn't allowed to take \infty - \infty.

Posted: **Sat May 03, 2008 9:52 am**

Jimmy_xor wrote:krauser2288 wrote:omg i was joking.......don't you read the whole post?!?!?!?!?

I read the whole post.. I can explain why I am going to report you.

The series you posted is logic mission 28 which you haven't completed yet. And now you posted it here so that someone "by accident" can help you with it.

what?!?!?!?are you listening what you are saying?i was joking!!!!!!!did it seems that i really need help for the mission?

omg i will delete it because you are capable to bann me for this

Posted: **Fri May 09, 2008 10:50 am**

Hay you can't divine each part with different value.

in this example x isn't the same with y

YOU SHOULD DIVINE ONLY WITH (X-Y) OR (X+Y) ORX OT Y OR Z WHI BELONGS TO IR

in this example x isn't the same with y

YOU SHOULD DIVINE ONLY WITH (X-Y) OR (X+Y) ORX OT Y OR Z WHI BELONGS TO IR

Posted: **Thu Jun 05, 2008 3:57 am**

Wow you guys are good at maths. Guess you all are High IQs.

Posted: **Fri Jun 06, 2008 12:07 pm**

yourmysin wrote:1$=100c=(10c)²=(0.1$)²=0.01$=1c

You cannot convert 10c to 0.1$ inside the ( )², you need to raise it to the ² power first, and then convert.

Based on the same logic:

60 hours = 3600 minutes = 60² minutes = 1² hours = 1 hour

Posted: **Sat Jun 07, 2008 3:26 am**

gotta square the units