Hello guys!

I've faced a minor difficulty in proving the statement "√3 is irrational number". My arguments are the following:

Let's suppose that √3 is rational then it can be expressed as p/q,  p,q∊Z which is irreducible, so √3=p/q <=> 3=p²/q² <=> p²=3q² 

And here is a problem from p²=3q² I concluded that p=3a, a∊Z after that the proof is led to contradiction =>

p/q can be reduced by 3. The ground for my doubt is we cannot conclude from product's divisibility the divisibility of it's factors

(maybe even these factors are the same) e.g. 12 | 3*8 but 12 ∤ 3 and 12 ∤ 8.

I apologize for my English)) and appreciate any attention.

 

 

p=3a leads to [latex]9a^2=3q^2\ or\ 3a^2=q^2[/latex], leading to in infinite chain, which cannot be.

Edited December 15, 20178 yr by mathematic
latex form

Not an expert, but I thought any root of a prime number is pretty much a definition of an irrational number.

28 minutes ago, druS said:

Not an expert, but I thought any root of a prime number is pretty much a definition of an irrational number.

Not the definition, no. It does need to be proven that given a prime p, then for any nonzero integers a, b that a^2/b^2 is not p.

5 hours ago, uncool said:

Not the definition, no. It does need to be proven that given a prime p, then for any nonzero integers a, b that a^2/b^2 is not p.

I said I wasn't an expert. I also did not say it was "the definition" but "pretty much a definition". This might not be math, but it is English. I guess that verbal qualifiers can be tricky things.

Out of interest, can you name a root of a prime that is rational? Genuine interest if the answer is yes.

 

13 hours ago, mathematic said:

p=3a leads to 9a2=3q2 or 3a2=q2 , leading to in infinite chain, which cannot be.

The point here is  9a^2=3q^2  leads to 3a^2=q^2  so q=3b, b ∈Z thus p/q=3a/3b which contradicts

with our assumption that p/q is irreducible.

  

 

Edited December 16, 20178 yr by vovka

6 hours ago, druS said:

I said I wasn't an expert. I also did not say it was "the definition" but "pretty much a definition". This might not be math, but it is English. I guess that verbal qualifiers can be tricky things.

Out of interest, can you name a root of a prime that is rational? Genuine interest if the answer is yes.

 

If a positive integer n is not a square, then its square root is irrational. This includes all primes. My point was that it isn't as simple as looking at the definition - it does take some facts about the integers in the first place, as vovka implicitly points out. 

druS said "Out of interest, can you name a root of a prime that is rational? Genuine interest if the answer is yes".

No, the roots of every prime number are all irrational.  But that is not exactly what you said before.

You said, before, "I thought any root of a prime number is pretty much a definition of an irrational number".  Now, I don't know what "pretty much" is intended to mean here but while it is true that "the root of every prime number is irrational", the converse "every irrational is the root of a prime number" is false.  $\pi$ and e are irrationals that are not roots of a prime number. 

You might be interested in this: Any number, rational or irrational, that satisfies an "n"th degree polynomial equation is "algebraic of degree n" (the rationals are all "algebraic of degree 1").  Numbers such as $\pi$ or e are NOT "algebraic" of any degree.  They are "transcendental" numbers.

On 12/17/2017 at 3:47 AM, uncool said:

If a positive integer n is not a square, then its square root is irrational. This includes all primes. My point was that it isn't as simple as looking at the definition - it does take some facts about the integers in the first place, as vovka implicitly points out. 

HallsofIvy

Sorry mate, I hadn't checked in as I simply thought it was a case of a specialist dumping on a basic novice. SO have to say I was very wrong in that assessment and thank you for your response.

FWIW I wasn't suggesting that all irrational numbers are prime square root, pie being an obvious example. In fact I would of thought that there are many, many irrational numbers, more than rational I guess.Yes I AM interested in the relationship of polynomials to transcendental numbers. But education must prevail for a while before it means much more to me than an abstract observation. I'll get there.

With persistence.

By the way, maybe someone here can answer a question I asked of a maths lecturer earlier this year but without clear response. Is plank's number irrational? I get that "plank-bar" would be due to the involvement of pie, but what about planks number itself? (apologies to Vovka for the hijack.)

Planck's "number" isn't really a number. It has units; your question is analogous to asking "Is 1 meter rational?" 

21 hours ago, uncool said:

Planck's "number" isn't really a number. It has units; your question is analogous to asking "Is 1 meter rational?" 

Haha!

I did realise that Planks constant was a unit, but I also thought that it was a an actual number. So, not the case.

 

Cheers

It is the part of the definition of the rational numbers that all the surds are the irrational numbers and √3 is a surd. That's why it is an irrational number. If you want to get further details about the rational and irrational numbers, then you can take a review of this post.....

 

link removed

1 hour ago, Juliana Sanabri said:

It is the part of the definition of the rational numbers that all the surds are the irrational numbers and √3 is a surd.

No it isn't.

The definition of irrational numbers is that they can't be written as a ratio of two integers.

The fact that surds are irrational is a consequence of that definition, but not part of it.