please try to solve these questions.
1a) What the heck is the definition of Q?
1b) Any even integer is of the form where n is some integer. (This is how an even number is defined.) Now square this:
Now, since n is an integer, so is . Again, since 2 and [tex]n^2 are integers so is . Thus is an integer it is also an even integer.
-Dan
Hello,
Topsquark.
In part(a) of question no 1 i have done this so far.
Q(a,b)= o if a<b.
Q(a-b,b)+1 if b is less than equal to a.
so
Q(3,4)=0 SINCE 3<4
GIVEN Q(a,b)= Q(a-b,b)+1) if b is less than equal to a
NOW
Q(14,3)=Q(11,3)+1
=Q(8,3)+1
=Q(5,3)+1
=Q(2,3)+1
SO AT THAT POINT 2 IS LESS THAN 3.
THEREFORE
=0+4
=4 IS THE ANSWER
To prove is irrational.
Assume,
Thus,
Thus,
.
Prime decompose .
Prime decompose .
Note,
has an even number of prime factors.
Note,
has an even number of prime factors.
But,
has an odd number of primes factors.
Thus, we have a contradiction because both sides are note equal by the uniquess part of the fundamental theorem of arithmetic.
This is a very general proof.
If p is a positive integer that is not a perfect square then is irrational.
Suppose that it is rational, where a & b are integers.
Let
By well ordering there is a first term in T, call it K.
Using the floor function and because p is not a square we have the following:
But is an integer less than K.
But is an integer which contradicts the minimumality of K.