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

[charmedbeauty]

Homework Statement

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

Homework Equations

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??

 

[Curious3141]

Not OK.

Homework Statement

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

Homework Equations

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

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?

 

[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.

 

[charmedbeauty]

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?

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.

??

 

[Curious3141]

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.

??

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

 

[charmedbeauty]

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

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.

 

[Curious3141]

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)

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

 

[charmedbeauty]

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

Ok thanks a bunch Curious!