Aarrgghh! (= please help)

[Pace]

Yup, more maths...(kind of urgent now)

Show that sqrt(n) is irrational if n is a positive integer that is not a perfect square.

(Note: Integer n is a perfect square if n = k^2 for some integer k. Hint: use a proof by contradiction, together with the Fundamental Theorem of Arithmetic)

Um, I'm lost, and I need to show this in ~2 hours

Thanks if you can help in time,

Paul.

[Raptor^]

Took a while to get this, but I think its sometihng along the lines of....

assume sqrt(n) is rational where n is not a perfect square.

so
n=k^2 for some non-integer, rational k

k is rational, so can be expressed as a fraction. Let k=a/b

so

n=(a/b)^2

express a and b in terms of their prime factors

n=(a1*a2*a3*..an/b1*b2*b3..bn)^2

According to FTA, there is a single unique product of primes for an integer.

so

n=(a1*a2*a3*..an * a1*a2*a3*..an)/(b1*b2*b3*..bn * b1*b2*b3*..bn)

n is an integer, so all the factors on the bottom of that must cancel with some on the top, so for all factors b1..bn there is a single, unique matching factor in a1..an.

so

a/b must be an int too, as it contains no factors that are not in (a/b)^2. However, a/b=k and k is non integer. This is a contradiction, so the converse must be true and sqrt(n) for n that is not a perfect square is irrational

Thats the gist of it, you might want to tidy and express it in more mathematical terms tho.

[The PIT]

Glad I don't study anymore!

[ayoub_ibrahim]

<font face="Verdana, Arial" size="2">Originally posted by The PIT:
Glad I don't study anymore!</font>

yes, but we're always learning