**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