# If n is a positive integer n then sqrt(4n-2) is irrational.

1. Apr 30, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

if n is a positive integer than √(4n-2) is irrational.

2. Relevant equations

3. The attempt at a solution

√(4n-2) Assume is rational

then by definition of rationality

√(4n-2)=p/q for some integers p,q where q≠0

so √(2(2n-1))=p/q by factoring out the 2 as common

√2 *√(2n-1) =p/q

but 2n-1 is always odd

so √(2n-1) is always odd

now let u=√(2n-1)

but √2*u cannot be factored since √2 is irrational and u is odd.

so √(4n-2)≠p/q

Therefore our assumption must have been wrong therefore

√(4n-2) must be irrational

Is this proof ok??

2. Apr 30, 2012

### Curious3141

Not OK.

Problem here. How do you know 2n-1 is a perfect square? Only integers can be meaningfully described as even or odd, so you're making an unfounded assumption that 2n-1 is a square.

There's a simple and instructive way to solve this by considering the cases where 2n-1 is a perfect square, and when it isn't, and, in the second case, the prime factorisation of 2n-1. However, you're probably more comfortable solving this in the typical mould you're used to.

So begin by stating your assumption that $\sqrt{4n-2} = \frac{p}{q}$, where p and q are coprime (no factors in common, because the fraction is reduced).

Now, $\frac{p^2}{q^2} = 4n - 2 \Rightarrow p^2 = 2(2n-1)q^2$.

and draw the obvious conclusion about p's evenness.

Now rearrange to: $p^2 + 2q^2 = 4nq^2$.

Clearly, the RHS is divisible by 4, which means the LHS also has to be divisible by 4. Since p is even, $4 | p^2$. Can you now derive a contradiction with regard to q?

3. Apr 30, 2012

### sweet springs

Hi.

You should look at even or odd character of p and q.
p/q is already reduced so not both p and q are even.
But you will see in another way that both p and q are even. Contradiction.

Regards.

Last edited: Apr 30, 2012
4. Apr 30, 2012

### charmedbeauty

Well as shown p2 and q2 both have a common factor of 2 but p2/q2 was already in it's most reduced form.

Contradiction, therefore √(4n-2) must be irrational.

??

5. Apr 30, 2012

### Curious3141

Yes, but how did you deduce that q is even?

6. Apr 30, 2012

### charmedbeauty

sorry, from the line

p2=2(2nq2-q2)

Since LHS is even therefore p is even.

ie p2/2 is also even

so p2/2=2nq2-q2

so p2/2=q(2n-1)

Since LHS is even then the RHS must be to.

clearly 2n-1 is not even

so obviously q has to be even if q(2n-1) is even.

7. Apr 30, 2012

### Curious3141

That should be q2(2n-1), but yes, that reasoning is sound.

8. Apr 30, 2012

### charmedbeauty

Ok thanks a bunch Curious!