Is sqrt(5) Rational? Understanding the Proof by Contradiction

[disregardthat]

It is not difficult to prove for arbitrary products of different primes (of course this will suffice for the general non-square case). The proof goes exactly the same way as the case for one prime. If b^2p_1...p_k = a^2, then p_1...p_k must be a factor of a, etc..

 

[TylerH]

Is it enough, once given that, if a|bⁿ, then aⁿ|b[supn[/sup], to say that:
n^1/2=p/q
n=p²/q²
nq²=p² Contradiction if n is not expressible as k², where k is an arbitrary integer.

The contradiction being that there is an uneven power of a prime factor on one side, which is impossible for a squared number.

 

[disregardthat]

Is it enough, once given that, if a|bⁿ, then aⁿ|b[supn[/sup], to say that:
n^1/2=p/q
n=p²/q²
nq²=p² Contradiction if n is not expressible as k², where k is an arbitrary integer.

The contradiction being that there is an uneven power of a prime factor on one side, which is impossible for a squared number.

I don't know what you mean on the top there (what is a?), but yes, this is what all the other proofs are driving at in more or less lengthy ways... It is merely a matter of comparing prime factorizations.

This method of proof generalizes neatly to the n'th root case. Comparing the exponents of the primes in the respective factorizations will yield the wanted contradiction.

 

[pmoseman]

How many times do we have to answer this guys question? Enough already!
We are not trying to do this to the nth root.
Unless you have a proof other than sqrt(a) = p/q just forget it. That's the only proof anyone has. The rest are just statements of fact.

 

[TylerH]

How many times do we have to answer this guys question? Enough already!
We are not trying to do this to the nth root.
Unless you have a proof other than sqrt(a) = p/q just forget it. That's the only proof anyone has. The rest are just statements of fact.

You don't have to at all.

 

[Gokul43201]

Are you calling me out on using the word "equations", instead of "expressions" (for which I provide solutions, so wtfc?)?

I'm saying your posts are more likely to be understood if you use standard terminology. If you don't care to have your posts understood, by all means, call an expression a banana.

It seems lackadaisical then for you to use terms such as "any" odd exponent, or a number "has" a square root.

I never used any such terms. But surely, you understand the difference between using inelegant English and wrong terminology. Not everyone here is a native speaker of English.

I could only understand the theorem you stated after seeing this: n = p_1^a \cdot p_2^b \cdots p_k^m (example: 12=2²3¹).

Sucks to be me, for taking the trouble of explaining a theorem to you (that someone else stated).

It can be stated clearly:
"The square root of a number is irrational, if the complete prime factorization of that number includes a prime any odd number of times."

Kudos to your English teachers.

I know the list of solutions to (1) (2) (3) (4) is probably hard to understand, but I simply did that to make a point.

If the list is hard to understand, then your point was probably missed.

Mark simply asked if it would be straighforward "...to show that two irrational numbers (a, b) have an irrational product if a and b are not equal, and a is not the reciprocal of b."

I feel like I have shown Mark's naive idea can be eked into a straightforward theorem.
"Two irrational numbers (X, Y) have an irrational product if rX and rY are not equal, and rX is not the reciprocal of rY." In fact, the only rational solution to most any of it, is 0, and 0 is a "rational" number only by acquaintance. In the most general way Mark was right on the money, although he might not be majoring in mathematics.

I mean, honestly, there is no justification to the approach of disproving his question by showing sqrt(24) * sqrt(6) = 12 because sqrt(24) and sqrt(6) are not really different irrational numbers, in the way 4 and 2 are not actually different numbers, they are different amounts of the same prime number, paying no mind to the number 1.

2 and 4 are not different numbers? Neither are sqrt(24) and sqrt(6)? That requires a serious twisting of either language or terminology.

He shouldn't have to worry about them being reciprocals either, like Matt Grime's example of pi *1/pi, since that would technically make a quotient, but alas. Are we going to clap Matt on the back for proving pi and pi^-1 are "different" irrationals that "produce" 1?

Like it or not, every number other than the multiplicative identity is indeed most definitely different from its inverse. You can not choose to use some non-standard interpretation of the word 'different' to argue that all of here are wrong.

 

[pmoseman]

Ok...

 

[Gib Z]

I wonder why this proof is not considered more standard:

Clearly the only integers whose square roots are integers are the perfect squares. So let's deal with a non perfect square, n. Suppose \sqrt{n} = \frac{p}{q} where we can suppose gcd(p,q) = 1, and q> 1. Then n = \frac{p^2}{q^2} which can't occur, since there are no common factors to cancel to get a denominator of 1 which is required to equal to integer, n. So we have a contradiction and square roots of non perfect squares are irrational.

 

[pmoseman]

The product of pi * 1/pi is no different from the quotient pi / pi and those are the exact same number.

sqrt(24) and sqrt(6) are different irrational numbers, but they are not different like 3 and 7 are different, they are different like 34 and 68. You can express 68 as 34 + 34, meaning it really is just two 34s. 34 and 34 and 34 are all the same number. sqrt (6) + sqr(6) = sqrt(24).

Mark was asking a question, his idea didn't need to get treated like that, and basically lied to.

If you read carefully I never said 2 and 4 are not different, I am telling you they are different quantities of the same number. When you multiply 7 * 2 then you are adding 7s together, and two is not added. sqrt6 and sqrt24 are the same numbers, if you pull the two, you are multipying the exact same number times the exact same number, and this is what Mark said made them rational.

I thought I was talking to same person who brought that theorem up in the first place, my mistake. I expected a response from them. I like the theorem and I liked your clarification.

Not all expressions are equations. We've established that and I am not about to call it a banana. I think the fact I supplied solutions to what I called equations is relevant, my English teacher is not relevant.

Missing my point is on you, my point is very pointy.

 

[disregardthat]

I wonder why this proof is not considered more standard:

Clearly the only integers whose square roots are integers are the perfect squares. So let's deal with a non perfect square, n. Suppose \sqrt{n} = \frac{p}{q} where we can suppose gcd(p,q) = 1, and q> 1. Then n = \frac{p^2}{q^2} which can't occur, since there are no common factors to cancel to get a denominator of 1 which is required to equal to integer, n. So we have a contradiction and square roots of non perfect squares are irrational.

It seems to me that every proof of this out there is more or less the same, but they may seem different by their long explanations as this is for many the first entrance into the world of mathematical proofs.

 

[mathwonk]

here's another explanation of gibz's proof which is actually also my favorite: if sqrt(n) = a/b is in lowest form, then n = a^2/b^2 is also in lowest form, but lowest form is unique and n/1 is a lowest form for n, so b^2 = 1 and a^2 = n. i.e. if sqrt(n) is rational, then sqrt(n) is an integer.