# Question re proof of irrational nos.

Hi everyone,

I've been reading about the proof of irrational nos. and I often encounter this phrase: "decreasing sequence of positive integers must be finite". What does this actually mean? Can anyone explain or point me to a link.

Here's a link of one proof I've read re proof of irrational nos: http://mathforum.org/library/drmath/view/55839.html

AKG
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A decreasing sequence of positive integers must be finite. Let s be any such sequence. Let s(1) denote the first element of the sequence. Well the sequence can then have at most s(1) elements, and if it had exactly s(1) elements, it would be the sequence s = (s(1), s(1)-1, ...., 2, 1). And s(1) is obviously a finite number.

So basically, in the context of proving irrational numbers, what that phrase means is that the numerator and denominator just gets smaller and smaller indefinitely (infinitely) and therefore (the numerator and denominator) cannot both be integers as required by rational numbers.

Is my conclusion right?

matt grime
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The phrase 'proving irrationals' doesn't mean anything at all.

That every set of positive integers has a least element is nothing to do with any proofs of any properties of that the irrationals may or may not have on its own.

I think the idea you're loooking for is reductio ad absurdum: show that if something is true it leads to an infinite strictly decreasing set of positive integers, which is impossible.

HallsofIvy
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I really have to ask: what do you mean by "proving rational numbers"??
Proving that rational numbers exist?? That is, proving that integers alone don't solve all equations that can be stated in terms of integers? What integer satisfies 2x= 1?

HallsofIvy said:
I really have to ask: what do you mean by "proving rational numbers"??
Proving that rational numbers exist?? That is, proving that integers alone don't solve all equations that can be stated in terms of integers? What integer satisfies 2x= 1?

It's actually proving the existence of irrational numbers. I'm asking what the phrase "decreasing sequence of positive integers must be finite" means in the context of proving the existence of irrational numbers. Please see link that I've posted.

matt grime said:
The phrase 'proving irrationals' doesn't mean anything at all.

What do you mean? I'm asking about "decreasing sequence of positive integers must be finite" not the phrase you just said.

matt grime said:
That every set of positive integers has a least element is nothing to do with any proofs of any properties of that the irrationals may or may not have on its own.

Isn't it that one proof of irrational numbers is that 2^(1/2) cannot be expressed as a ration of integers i.e. a/b because as the proof in the link I posted has shown, both a and b decreases indefinitely (infinitely) and therefore cannot be integers because "decreasing sequence of positive integers must be finite".

matt grime said:
I think the idea you're loooking for is reductio ad absurdum: show that if something is true it leads to an infinite strictly decreasing set of positive integers, which is impossible.

Yup. Like I've mentioned above.

Have I understood the proof right?

Please correct me if I haven't. Thanks!

matt grime
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The phrase 'the proof of irrationals nos' appears in post one. There is no such things as 'the proof of irrational nos'. Some words are missing. In post three you use the phrase 'in the context of proving irrationals'.

You might have meant

1. prove that irrational numbers exist
2. prove that some specific number is irrational.

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HallsofIvy
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The set of positive integers is "well ordered"- that is, any set of positive integers contains a smallest member. In particular, "1" is the smallest of all positive integers so if {an} is a decreasing sequence of positive integers, then a0 is the largest number in the sequence and the sequence cannot have more than a0[\sub] members. That's what AKG said in the first response to your post.

The method of proving that $\sqrt{2}$ is not rational is called "infinite descent". Assuming that $\sqrt{2}= \frac{m}{n}$ we can show that both m and n must be even: m= 2a, n= 2b so that $\sqrt{2}= \frac{a}{b}$. Of course, a< m, b< n and we can keep doing that- we get an infinite sequence of decreasing positive integers which is impossible.

Often that hidden by simply asserting that any rational number can be written in "lowest terms"- that m and n have no common factors. Of course, then, showing that both m and n are even is an immediate contradiction.

A neat, simple proof of the irrationality of phi follows from it's geometric definition, the ratio A/B that divides a length A+B into lengths A and B such that (A+B)/A = A/B. If one takes (A+B)/A to be in lowest terms, then A/B is in even lower terms, which is a contradiction.

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StatusX
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If you just want to prove irrational numbers exist, probably the simplest examples are logs. For example, log3 5 must be irrational because if it were equal to m/n for some positive integers m and n, then 3m/n=5, so 3m=5n, which is clearly absurd (for example, one is divisible by 5 and the other isn't).

matt grime said:
The phrase 'the proof of irrationals nos' appears in post one. There is no such things as 'the proof of irrational nos'. Some words are missing. In post three you use the phrase 'in the context of proving irrationals'.

You might have meant

1. prove that irrational numbers exist
2. prove that some specific number is irrational.

Yes. What I meant was #1

In that case, perhaps you would be interested in the concept of a "Dedekind cut".

http://en.wikipedia.org/wiki/Dedekind_cut

You might also pick up a book on Number Theory -- I recommend Hardy and Wright's "Introduction to the Theory of Numbers", but almost all introductory texts cover irrational numbers.

By the way, the proof you originally gave in post 1 doesn't actually prove the existence of irrationals -- for example, it could simply be that sqrt(2) isn't actually a number. It is in fact impossible to "prove" the existence of irrational numbers -- the best you can do is show that there exists some quantities that cannot be expressed as a rational numbers. That's not a proof.

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gnomedt said:
In that case, perhaps you would be interested in the concept of a "Dedekind cut".

http://en.wikipedia.org/wiki/Dedekind_cut

You might also pick up a book on Number Theory -- I recommend Hardy and Wright's "Introduction to the Theory of Numbers", but almost all introductory texts cover irrational numbers.

Thanks. I'll check that out.

gnomedt said:
By the way, the proof you originally gave in post 1 doesn't actually prove the existence of irrationals -- for example, it could simply be that sqrt(2) isn't actually a number. It is in fact impossible to "prove" the existence of irrational numbers -- the best you can do is show that there exists some quantities that cannot be expressed as a rational numbers. That's not a proof.

Technically you're right. You cannot prove irrational numbers but you can prove that there are numbers that are not rational (i.e. cannot be expressed as a ratio of integers) and therefore are irrational.

Thanks again

matt grime
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"Prove irrationals exist" not "prove irrationals". Sorry to bang on about this, but please try to make your sentences correct, especially after people have pointed out the mistake at length several times - maths is all about clarity and communication.

matt grime said:
"Prove irrationals exist" not "prove irrationals". Sorry to bang on about this, but please try to make your sentences correct, especially after people have pointed out the mistake at length several times - maths is all about clarity and communication.

Yup. Thanks for pointing that out. I had "Prove irrationals exist" in mind but expressed it incorrectly. 'Will be careful next time.

:)