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Dido525 Group TitleBest ResponseYou've already chosen the best response.1
dw:1358647793062:dw
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Well since root 2 is irrational, an irrational number raised to an irrational number should still be irrational.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
But that's not always true. Hmm... Intresting. One moment.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Thing is, this is NOT an irrational number. It's rational.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Use proof by contradiction.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Assume is IS rational.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
ok go on
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
i dont have a very nice method but i guess this will help you let's say \[x= 2 ^{\sqrt2}\] take log on both the sides \[\log x = \sqrt {2}\log2\] find the value of right side and finally \[X= e ^{\sqrt2 \log 2}\] and i am sure that would be irrational
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
dw:1358648074024:dw
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
you need to use calculator
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
dw:1358648102171:dw
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
dw:1358648120754:dw
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
This is obviously rational.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
^ two.. not 4
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Right. Still rational though.
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
there is a fallacy as far as i can see in your solution that you have missed square root on the exponent @djdo525
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
sorry its Dido
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
ok guys one correction, that root 2 power root 2 is rational . Sorry But prove that root 2 power root 2 is transcendental
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Also, PLEASE check whether it is transcendental or not , I am not sure. But I need your help.
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
lol that was horrible
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
OK friends I am totally confused. http://www.math.hmc.edu/funfacts/ffiles/30002.35.shtml Here I got that root 2 power root 2 is irrational..
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
^ and as well as transcendental
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.1
@Dido525 , your example is incorrect, you showed (sqrt2)^(sqrt2)^sqrt2 is rational but sqrt2^sqrt2 is irrational
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Yeah. That's what he wanted.
 one year ago
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