Hi
... ,
Question 1:
i'm not sure if you still need the solution to your first question of irrationality, but here it is anyways:
let 'log' from here on denote "logarithm of base 10" :
log(10) = 1 ---> log(15) > 1
ie. log(15) = a/b ; where 'a' & 'b' are
mutually prime integers, and where a > b
---> 10^(a/b) = 15 ---> 10^a = 15^b ;
since a > b, then let a = b + k ; where 'k' is
integral.
ie. 10^b.10^k = 15^b ----->
10^k = (3/2)^b
LHS = 10^k is always
integral for all integer 'k'
but, (3/2) is non-integral and can't be reduced/simplified further, and since 'b' is integral, then
RHS = (3/2)^b is always
non-integral for all integer 'b'.
hence, we arrive at the contradiction that although
LHS = RHS, no numbers in the real system can be both integral and non-integral simultaneously.
ie. initial assumption of rationality is refuted.
Question 2:
As to your second question of
Greatest Common Divisor, there is nothing wrong with what
turtle_2468 has done, except that his method is a bit unnecessarily lengthy for my liking... so here's what i would have done:
consider the ratio {or fraction}:
(10!15!)/(5!20!)
the (5!) in the denominator
cancels completely against (10!), and the (15!) in the numerator also
cancels completely against (20!) to give:
(10!15!)/(5!20!) = (6.7.8.9.10)/(16.17.18.19.20)
which then easily cancels down again to: (6.7.8.9.10)/(16.17.18.19.20) =
(3.7)/(4.17.19)
this does not reduce further, hence the GCD is very simply:
GCD{10!15!, 5!20!} =
(10!15!)/(3.7) =
(5!20!)/(4.17.19)
hope that helps
P.S. personally, if you encounter this type of question in an exam, i'd advise you not to use the technique of decomposing into prime factors and adding up the exponents one by one as
turtle_2468 suggested ... since it's very easy to make a mistake that way. But cancelling out and reduction using fractions is much easier and more expedient.
