Very easy Maths Question~~(30)

Method of proving 3^1/2 (root 3) as an irrational number:

Let √3=p/q, which p and q are natural number and (p,q)=1 (p and q's H.C.F. is 1)

(√3)^2=(p/q)^2

3=p^2/q^2

p^2=3q^2----------(1)

From (1), 3∣p (p can be divided by 3)

Let p=3m, and substitute it into (1),

(3m)^2=3q^2

9m^2=3q^2

3m^2=q^2-------------(2)

From (2), 3∣q (q can be divided by 3)

Thus, (p,q)≠1 (since both p and q can be divided by 3),
contradicted with (p,q)=1

Therefore, since √3 can not be written in the form of
p/q, it is an irrational number.

General method to prove irrational number:

Let √y = p/q, p and q are natural number, and (p,q)=1
(the H.C.F. of p and q is 1)

( √y)^2= ( p/q)^2

y=p^2/q^2

q^2y=p^2--------------(1)

From (1), y∣p (p can be divided by y)

Let p=my, and substitute it into (1),

q^2y=(my)^2

q^2y=m^2y^2

q^2=m^2y-----------------(2)

From (2), y∣q (q can be divied by y)

Thus, (p,q)≠1, (since both p and q can divided by y), contradicied
with (p,q)=1

Since √y can not be written in the form of p/q, it is an irrational
number

Hope this can help you~

Assume the contrary, assume √3 is rational.

Hence, we can express √3 = p/q, where p and q are relatively prime, p, q are integers.

Squaring both sides, we get 3 = p^2/q^2,

Hence, 3q^2 = p^2, where p^2 is a multiple of 3.

Then, it implies p is a multiple of √3.

2009-06-01 20:20:25 補充：
Due to this argument, we let p = a√3, where a is some integer

So, 3q2 = (a√3)2

q = +- a

2009-06-01 20:20:37 補充：
As a result, p and q have the common factor a, which contradicts the assumption that p and q are relatively prime.

Therefore, √3 is irrational.

P.S. This is a general method to prove a sqrt is an irrational number. (Contradiction)

不識睇呀~~~~~~~ =.=