Talk Nerdy To Me ;)
Post Reply
 
Thread Rating:
  • 0 Votes - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
20-12-2016, 02:05 PM
RE: Talk Nerdy To Me ;)
(19-12-2016 11:12 AM)jennybee Wrote:  If you pile it on top of your head, it's a yoga bun Wink

Yoga buns on top of my head? Consider

Yes. Yes, please! Yes

Find all posts by this user
Like Post Quote this message in a reply
[+] 3 users Like TheGulegon's post
20-12-2016, 02:09 PM
RE: Talk Nerdy To Me ;)
Can a cat be nerdy? Let's find out, shall we????

[Image: 1e945a0e164d5114092c3b7324feef7a.jpg]

[Image: Nerdy-Cat-Names_Cat-Names-City_Geeky-Cat...=300%2C201]

[Image: live-long-and-pawspurr-cat.jpg][/b]

Well, it looks like I've proven that not all cats are cool cats....some are nerdy cats. Yes

Shakespeare's Comedy of Errors.... on Donald J. Trump:

He is deformed, crooked, old, and sere,
Ill-fac’d, worse bodied, shapeless every where;
Vicious, ungentle, foolish, blunt, unkind,
Stigmatical in making, worse in mind.
Find all posts by this user
Like Post Quote this message in a reply
[+] 6 users Like dancefortwo's post
20-12-2016, 02:20 PM
RE: Talk Nerdy To Me ;)
(20-12-2016 02:09 PM)dancefortwo Wrote:  Can a cat be nerdy? Let's find out, shall we????

[Image: 1e945a0e164d5114092c3b7324feef7a.jpg]

[Image: Nerdy-Cat-Names_Cat-Names-City_Geeky-Cat...=300%2C201]

[Image: live-long-and-pawspurr-cat.jpg][/b]

Well, it looks like I've proven that not all cats are cool cats....some are nerdy cats. Yes

I think that last one is Kernel's cat Tongue
Find all posts by this user
Like Post Quote this message in a reply
[+] 4 users Like jennybee's post
20-12-2016, 05:51 PM
RE: Talk Nerdy To Me ;)
- What do you do with a dead chemist?
- Barium

"Freedom is the freedom to say that 2+2=4" - George Orwell (in 1984)
Find all posts by this user
Like Post Quote this message in a reply
[+] 5 users Like Leerob's post
20-12-2016, 07:32 PM
RE: Talk Nerdy To Me ;)
(20-12-2016 02:09 PM)dancefortwo Wrote:  Can a cat be nerdy? Let's find out, shall we????

[Image: live-long-and-pawspurr-cat.jpg][/b]

Well, it looks like I've proven that not all cats are cool cats....some are nerdy cats. Yes

Oh, hell no, that cat is the demon seed, and someone's arm is about to have parallel lines of ripped flesh in a few seconds.
Find all posts by this user
Like Post Quote this message in a reply
20-12-2016, 08:46 PM
RE: Talk Nerdy To Me ;)
*whispers picard is better than kirk

[Image: Guilmon-41189.gif] https://www.youtube.com/channel/UCOW_Ioi2wtuPa88FvBmnBgQ my youtube
Find all posts by this user
Like Post Quote this message in a reply
[+] 2 users Like Metazoa Zeke's post
08-01-2017, 12:15 PM
RE: Talk Nerdy To Me ;)
[Image: zZFnDuO.gif]

Don't Live each day like it's your last. Live each day like you have 541 days after that one where every choice you make will have lasting implications to you and the world around you. ~ Tim Minchin
Find all posts by this user
Like Post Quote this message in a reply
[+] 3 users Like Commonsensei's post
08-01-2017, 12:47 PM
RE: Talk Nerdy To Me ;)
Don't worry. As long as you hit that wire with the connecting hook at precisely eighty-eight miles per hour the instant the lightning strikes the tower... everything will be fine.

There is only one really serious philosophical question, and that is suicide. -Camus
Find all posts by this user
Like Post Quote this message in a reply
[+] 1 user Likes GirlyMan's post
08-01-2017, 02:35 PM (This post was last modified: 08-01-2017 02:53 PM by Kernel Sohcahtoa.)
RE: Talk Nerdy To Me ;)
Let G be a group with the general operation *; thus, we can denote this as <G,*>. By the definition of a group, the elements in <G,*> will be associative [a*(b*c)=(a*b)*c.] and will possess an identity element e (for example, a*e=a and e*a=a) along with an inverse (for example, a*a'=e and a'*a=e).

Now, IMO, here's a pretty cool fact: if G is a group and a,b are elements in G, then (ab)^-1 (or the inverse of ab) is equal to b^-1a^-1 (or the inverse of b times the inverse of a). Now, let the operation in G be multiplicative. To prove this, multiply the left side of b^-1a^-1 by ab. Thus, ab(b^-1a^-1)
=a(bb^-1)a^-1 (via the associative law of algebra)
=aea^-1 (note that bb^-1= e)
=aa^-1 (note ae=a)
=e.

Since the product of ab and b^-1a^-1 resulted in the identity element e, then this means that ab and b^-1a^-1 are inverses of each other. Consequently, the inverse of ab is b^-1a^-1
or (ab)^-1=b^-1a^-1.

"I'm fearful when I see people substituting fear for reason." Klaatu, from The Day The Earth Stood Still (1951)
Find all posts by this user
Like Post Quote this message in a reply
[+] 2 users Like Kernel Sohcahtoa's post
08-01-2017, 02:57 PM (This post was last modified: 08-01-2017 03:35 PM by jennybee.)
RE: Talk Nerdy To Me ;)
(08-01-2017 02:35 PM)Kernel Sohcahtoa Wrote:  Let G be a group with the general operation *; thus, we can denote this as <G,*>. By the definition of a group, the elements in <G,*> will be associative [a*(b*c)=(a*b)*c.] and will possess an identity element e (for example, a*e=a and e*a=a) along with an inverse (for example, a*a'=e and a'*a=e).

Now, IMO, here's a pretty cool fact: if G is a group and a,b are elements in G, then (ab)^-1 (or the inverse of ab) is equal to b^-1a^-1 (or the inverse of b times the inverse of a). Now, let the operation in G be multiplicative. To prove this, multiply the left side of b^-1a^-1 by ab. Thus, ab(b^-1a^-1)
=a(bb^-1)a^-1 (via the associative law of algebra)
=aea^-1 (note that bb^-1= e)
=aa^-1 (note ae=a)
=e.

Since the product of ab and b^-1a^-1 resulted in the identity element e, then this means that ab and b^-1a^-1 are inverses of each other. Consequently, the inverse of ab is b^-1a^-1
or (ab)^-1=b^-1a^-1.

Tongue

[Image: 9596429.gif]
Find all posts by this user
Like Post Quote this message in a reply
[+] 6 users Like jennybee's post
Post Reply
Forum Jump: