Proving a number is irrational

[Miike012]

I picked up a book by Stephen Abbott called "Understanding Analysis" and it begins talking about rational and irrational numbers then it goes on proving how √2 is irrational. The proof is easy to understand but I wanted to use the same exact proof on a number I knew was rational.

Let (p/q)² be a rational number equal to 4 where p and q have no common factors and q ≠0.

(p/q)² = 4 which implies p² = 4*q². From this we can see that p² is a multiple of 4 and hence so is p. Let p = 4a where a is an integer.

16a² = 4*q² , 4a² = q² which implies q² is a multiple of 4 and hence q is also.


Question:
There for we now know the ratio p/q is not in lowest terms, it has a common factor of 4.

So how am i suppose to interpret this? This is the same exact proof that proved √2 is irrational by showing p and q have common factors. Well I just showed you that √4 = p/q where p and q have common factors but √4 = 2 is obviously rational.

 

[gopher_p]

... From this we can see that p² is a multiple of 4 and hence so is p.

This is where you went wrong. While it is true that, for any integer p, 2|p whenever 2|p^2, this is due to 2 being prime. 4 is not prime.

 

[Dustinsfl]

I picked up a book by Stephen Abbott called "Understanding Analysis" and it begins talking about rational and irrational numbers then it goes on proving how √2 is irrational. The proof is easy to understand but I wanted to use the same exact proof on a number I knew was rational.

Let (p/q)² be a rational number equal to 4 where p and q have no common factors and q ≠0.

(p/q)² = 4 which implies p² = 4*q². From this we can see that p² is a multiple of 4 and hence so is p. Let p = 4a where a is an integer.

16a² = 4*q² , 4a² = q² which implies q² is a multiple of 4 and hence q is also.Question:
There for we now know the ratio p/q is not in lowest terms, it has a common factor of 4.

So how am i suppose to interpret this? This is the same exact proof that proved √2 is irrational by showing p and q have common factors. Well I just showed you that √4 = p/q where p and q have common factors but √4 = 2 is obviously rational.

The ratio for the proof is always in lowest terms. To start with 2/4 instead of 1/2 is nonsensical.

As a caveat, I believe the proof in most books always states $\text{gcd}(p,q) =1$ or says they are relatively prime.

 

[jgens]

You cannot conclude that p is a multiple of 4. You can only conclude that is a multiple of 2. This avoids the contradiction.

 

[Miike012]

Revised Solution:

(p/q)² = 4 which implies p² = 4*q2. From this we can see that p² is a multiple of 4 and p a multiple of 2. Let p = 2a where a is an integer.


p² = 4*q² or 4a² = 4*q². From this equation we can see

Equation 1: (a^2 = q².)

Is this correct?
QUESTIONS: If we know p is a multiple of 2 and from equation 1 we see that a^2 = q² where a is any integer, from this equation how can we conclude from eq 1 that q is equal to 1?

or is it the fact that p and q do not have common factors which tells us that the ratio p/q is reduced, assuming p =/= q and that alone implies that there is a rational number when squared that is equal to 4. Is that correct?

 

[Office_Shredder]

QUESTIONS: If we know p is a multiple of 2 and from equation 1 we see that a^2 = q2 where a is any integer, from this equation how can we conclude from eq 1 that q is equal to 1?

The fact that p²=4q² and p and q have no common factors means that the only thing that can divide p are multiples of 2. From there the fact that p and q have no common factors implies that p must be equal to 2 (otherwise q would have to be divisible by 2 as well)

or is it the fact that p and q do not have common factors which tells us that the ratio p/q is reduced, assuming p =/= q and that alone implies that there is a rational number when squared that is equal to 4. Is that correct?

You're trying to do a proof by contradiction and failing to find a contradiction (since there isn't one to be found), so you shouldn't expect to prove anything like this. You can get around to calculating what the square root of 4 is using what I typed above but it really relies on the fact that 2²=4 to begin with, so I don't think there's any value in thinking about it too much. It's better to recognize that trying to prove by contradiction that 4 has no square root won't get you very far because it's not true that 4 has not square root (and figure out where the hole is in your calculation so you can throw it away with a peaceful mind) than to try to figure out how you can salvage the calculations you made and actually prove something