you dint even mention that the solutions you posted were not yours! 

but he also did not say that they were his

did he?

forget that and try 2 solve the questions

thats what they are mant for

ans 5)

another solution can be

let x=positive rational number

(p/q)^p/q                      where GCD (p,q)=1      and p and q are integers and q not equal to 0,1 q>p

this means  (p/q)^p  never comes out of the root as an integer          thus (p/q)^{p is irrational}

also (p^p/q) / q^p/q 

here q^p/q is irrational 

to make it rational, only option is q=1

but then (p^p/q) / q^p/q becomes p^p/1  which is an integer

also earlier we said q not equal to 1

here a contradiction arises

hence for x^x to become rational x has to be an integer

Q 6) x^2 = x+x+x+x+....x times is only valid for positve integral values....... so its a discountinous function.....

hence cant differentiate.....

n ya i wanted to ask u tat Q 1) is one of the lettr a b c d e f g h  a decimal point.....???????

Q 5) see let us assume x to be a fraction in the form of p/q.... k nw x^x = (p/q)^p/q......

nw it can be read as p/q to the power of p under the root of q....( i hope u get me).........

as p/q is the least reduced fraction rite..... hence we can rite p as k*q*y wer k is a constant n y is any fractional no less than one........ so u can take the root of q away by nullifying the power p= kqy now remains only sum power of q which is an integer n y the fraction..... a fraction raised to a no less than 1 is irrational.............