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Proof of irrationality of sqrt(2)

by StatusX
Tags: irrationality, proof, sqrt2
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StatusX
#1
Aug25-06, 05:52 PM
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Here'a a really simple and intuitive proof that sqrt(2) is irrational, which extends immediately to the nth root of any rational number that isn't a perfect nth power.

It goes like this. Let r be a rational number, and let a, b be integers such that r=a/b and (a,b)=1 (so that a/b is r in lowest terms). Then clearly r2=a2/b2, and it is clear that (a2,b2)=1, so that this is r2 in lowest terms. Thus we see that the square of any rational number, written in lowest terms, has a perfect square in both the numerator and denominator. So we can conclude that if a rational number does not have this property, it is not the square of a rational number, and so its square root is irrational. Since 2 is not a perfect square, sqrt(2) is irrational.

This seems a lot more straightforward than the standard proof by contradiction. Is this well known, and why isn't it more popular?
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matt grime
#2
Aug25-06, 06:18 PM
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Yes this is well known though it is more commonly written along the lines of clearing denominators and looking at prime factors: you are using uniqueness of prime decomposition (or that the representation of an rational number in lowest terms is unique, and as this is usually not proved before proving the irrationality of sqrt(2)).
mathwonk
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Aug25-06, 10:33 PM
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i thjink i posted this proof here a couple of years ago.it is my favorite since i thought of it a decade or so ago while teaching the course.

i.e.if a/b is a square root of 2 in lowest etrms then a^2/b^2 is 2/1 in lowest etrms, so a^2 = 2, but this is not true.

mathwonk
#4
Aug25-06, 10:39 PM
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Proof of irrationality of sqrt(2)

see page 7 of my webpage elementary algebra notes:

4000.10-13.pdf rational numbers, irrational numbers, rational roots theorem, real numbers, infinite decimals, geometric series and repeating decimals, adjoining square roots to Q
HallsofIvy
#5
Aug26-06, 05:17 AM
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It is essentially Euclid's proof- except that Euclid proved the "it is clear that (a2,b2)=1, so that this is r2 in lowest terms" part rather than just saying it.
StatusX
#6
Aug26-06, 10:04 AM
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This isn't a proof by contradiction. I think it's more intuitive, even if the details are a little harder. Like proving (a2,b2)=1. This is "obvious", but the simplest way I can think to prove it is using prime decomposition. So Euclid's is probably nicer, if less satisfying.
Werg22
#7
Aug26-06, 10:53 AM
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Your proof however does not prove the existence of a irrational limit of the function x^1/2 when x - > 2. Your proof either means the square root is irrational or that it is inexistent.
matt grime
#8
Aug26-06, 05:32 PM
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Ho, hum, since we all agree that the square root of two exists, as much as anything else exists, this is a pointless observation.
Werg22
#9
Aug26-06, 10:24 PM
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Quote Quote by matt grime
Ho, hum, since we all agree that the square root of two exists, as much as anything else exists, this is a pointless observation.
Yes but still essential for the proof to be rigorous...
matt grime
#10
Aug27-06, 03:14 AM
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No, it isn't. The proof is rigorous. If there is a symbol in some extension of Q whose square is 2 that number is not in Q. We don't need for there to be a symbol whose square is two at all. As a matter of fact there is an extension where there is a symbol whose square is 2, but that is neither here nor there for this proof.

The existence or otherwise doesn't alter the rigor of the proof, but it does affect the 'point' of the proof.
raoulh
#11
Aug30-06, 06:02 AM
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I can't follow this proof as I don't have a clue what (a,b)=1 means (I've just finished GCSEs and am about to do A levels if anyone familiar with the english school system wants to know what level I'm at). I've encountered and can easily understand the normal proof as it just relies upon basic algebra. I don't see how this proof is "simpler".
matt grime
#12
Aug30-06, 06:26 AM
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(a,b) means the highest common factor of a and b: the largest positive number that divides them both. So (a,b)=1 just means that the fraction a/b is in lowest terms.

The point is it is not a needless proof by contradiction and involves fewer steps. That makes it simpler. Simpler in terms of proof complexity does not mean simpler for the reader to understand, necessarily: it depends upon the underlying knowledge of the reader. Perhaps 'cleaner' would be a better description of it to avoid any ambiguity.

If you do maths at university for instance you will meet the following theorem or a variation of it: any continuous function from the interval [0,1] to the reals is bounded and attains its bounds. The 'easy' proof in terms of material required to understand the proof is messy. The 'hard' proof is much actually much simpler since it is a simple consequence of something called the Heine-Borel theorem and the fact that the continuous image of a compact space is compact, both of which are very easy to prove once you know what they are.

Finally, here is an incredibly simple proof that no n'th root of 2 is a rational number for n greater than or equal to 3: suppose that 2 = (a/b)^n for a and b integers, then 2b^n = a^n, or b^n+b^n=a^n, but this contradicts Fermat's Last Theorem (which we know to be true and incredibly complicated to prove).
raoulh
#13
Aug30-06, 07:10 AM
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OK, thanks for the explanation. I think the use of the word "simple" was too ambiguous though in the original post.
HallsofIvy
#14
Aug30-06, 10:33 AM
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Well, you have to understand that mathematicians use the term "simple" to mean "I guess I could work this out if I had enough time".

Another useful phrase is "intuitively obvious to the most casual observer" which means "I have no idea how to do this so I'm just going to fake it"!

I really did once have a teacher who worked out a proof on the board, said "Now it is easy to see that..." and suddenly stopped and said "Now, why is that easy?". He then sat down in front of the class, wrote on a paper without saying anything for several minutes, then stood up and said "Yes, it is easy!".
raoulh
#15
Aug30-06, 10:42 AM
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That reminds me of some definitions I found on a joke site:

Clearly: I don't want to write down all the "in-between" steps.
Trivial: If I have to show you how to do this, you're in the wrong class.
It can easily be shown: No more than four hours are needed to prove it.
Check for yourself: This is the boring part of the proof, so you can do it on your own time.
Hint: The hardest of several possible ways to do a proof.
Brute force: Four special cases, three counting arguments and two long inductions.
Elegant proof: Requires no previous knowledge of the subject matter and is less than ten lines long.
Similarly: At least one line of the proof of this case is the same as before.
Two line proof: I'll leave out everything but the conclusion, you can't question 'em if you can't see 'em.
Briefly: I'm running out of time, so I'll just write and talk faster.
Proceed formally: Manipulate symbols by the rules without any hint of their true meaning.
Proof omitted: Trust me, It's true.


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