Prove that any positive rational number q is uniquely represented as q=p1^r1…pk^rk

Question: Prove that any positive rational number q is uniquely represented as q=p1^r1…pk^rk, where p1…pk are prime numbers, r1…rk are integer numbers (+ or -).

1b: Prove that sqrt(q) is rational number if and only if the numbers r1…rk are even.

1c: Let t=2cos(72degrees), prove that t^2+t-1=0.

Solution:

1a) Given
q=p1^r1.p2^r2.p3^r3…pk^rk
let us consider
q=12
prime factors are 2*2*3
So,here p1=2,r1=2
p2=3,r2=1
Here we can represent a positive rational
in the form of q=p1^r1.p2^r2.p3^r3…pk^rk
which is 12=2^2 * 3^1.
Hence we can say that any positive rational number q is uniquely represented as q=p1^r1…pk^rk, where p1…pk are prime numbers, r1…rk are integer numbers (+ or -).


1b) From the above example
we have assumed q=12 which is a rational
and if we take square root of it.

√12=2*√3
       = 3.46410162..
       which is a irrational number.
As r1=2(even), but r2=1 which is odd, the sqrt has become irrational number


Similarly we can take a number 16
q=16
p1=2,r1=4(even)

Here r1 is even, so does the √q=√16=4 which is rational number
Here we can conclude that, √q is rational number if and only if the numbers r1…rk are even.

1c)
t=2cos(72) cos in degrees
=2*0.309
=0.6180
=0.62

t^2+t-1=0
substitute value of t in the above equation

=0.62^2 +0.62 -1
=0.38+0.62-1
=1-1
=0
hence proved

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