So...Is Garnate the new GB moderator?

ah, okay, i see now
Byron was a man with many hats, he wasn't just Noor's awesome lover.
 
ah, okay, i see now
Byron was a man with many hats, he wasn't just Noor's awesome lover.

Oh yes, many hats on Lit, online and offline. Although he pretty much told me all he was doing, I did not realize the extent of it until after his death. I seriously don't know where he found the time for it all, especially considering how much time he spent with me.
 
Oh yes, many hats on Lit, online and offline. Although he pretty much told me all he was doing, I did not realize the extent of it until after his death. I seriously don't know where he found the time for it all, especially considering how much time he spent with me.

I feel honored to have learned some of his story. I can tell he was loved by many.
 
I feel honored to have learned some of his story. I can tell he was loved by many.

Yeah...yeah. Back to Garnate. Could you PM her and tell her we need a clean up in Aisle 5, spammer shot some spooey on the linoleum.
 
Oh yes, many hats on Lit, online and offline. Although he pretty much told me all he was doing, I did not realize the extent of it until after his death. I seriously don't know where he found the time for it all, especially considering how much time he spent with me.

A clever and quick man, best in the impromptu.
 
Prove it...

Either \sqrt{2}^{\sqrt{2}} is a rational number and we are done (take a=b=\sqrt{2}), or \sqrt{2}^{\sqrt{2}} is irrational so we can write a=\sqrt{2}^{\sqrt{2}} and b=\sqrt{2}. This then gives \left (\sqrt{2}^{\sqrt{2}}\right )^{\sqrt{2}}=\sqrt{2}^{2}=2, which is thus a rational of the form a^b.
 
Either \sqrt{2}^{\sqrt{2}} is a rational number and we are done (take a=b=\sqrt{2}), or \sqrt{2}^{\sqrt{2}} is irrational so we can write a=\sqrt{2}^{\sqrt{2}} and b=\sqrt{2}. This then gives \left (\sqrt{2}^{\sqrt{2}}\right )^{\sqrt{2}}=\sqrt{2}^{2}=2, which is thus a rational of the form a^b.

syntax error!
 
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