Root 2 is irrational

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A rational number is a number that can be written as ab \frac{a}{b}, where aa and bb are integers.

An irrational number is a real number that is not rational.

How do we know that irrational numbers exists? Back during Pythagoras' time, they did not believe that there were irrational numbers (they were being irrational!) and you could be drowned for believing something that was true!

Let's show that 2 \sqrt{2} is irrational:

Proof: We will prove this by contradiction. Suppose that 2 \sqrt{2} is not irrational. Then it is rational, and hence is of the form 2=ab\sqrt{ 2} = \frac{a}{b} . We may make the assumption that the fraction is in the lowest terms, or that aa and bb have no common factors.

Squaring the equation, we get that a2=2b2 a^2 = 2 b^2 . Since a2a^2 is even, hence aa is even. Let a=2x a = 2x , where xx is an integer. Substituting this in, we get 4x2=2b2 4x^2 = 2b^2 , or that 2x2=b2 2x^2 = b^2 . We repeat the above argument. Since b2b^2 is even, thus bb is even.

But this says that aa and bb are both even, and hence are both divisible by 2. This contradicts the original assumption that aa and bb have no common factors! _\square

Thus we know that 2 \sqrt{2} is irrational, and thanks to the advance of mathematics, we do not need to sacrifice our lives.

How do you show that 3 \sqrt{3} is irrational? What about 4 \sqrt{4} ?

Note by Chung Kevin
5 years, 8 months ago

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If I may be pedantic, you meant to say, "An irrational number is a REAL number that is not a rational number."

I think a much better proof goes by using much more generality.

Theorem: If pp is prime, then p\sqrt{p} is irrational.

Proof: Suppose, by way of contradiction, that p=ab\sqrt{p}=\frac{a}{b}, where a,bZa,b\in\mathbb{Z} are coprime. Thus, p=a2b2p=\frac{a^2}{b^2}. Since pZp\in\mathbb{Z} and a2a^2 and b2b^2 are coprime, we must have that b=1b=1, so p=a2p=a^2. But, since pp is prime, we have pa2p\neq a^2 for all aZa\in\mathbb{Z}. Contradiction. \boxed{}

Because 2 is prime, it follows that 2\sqrt{2} is irrational.

Jacob Erickson - 5 years, 8 months ago

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It's much simpler to consider the polynomial x22=0x^2-2=0. By the Rational Root Theorem, all rational roots of this polynomial must be either 1, 2, -1, or -2. None of these satisfy the equation, therefore there are no rational solutions to x=2x=\sqrt{2}

Edward Jiang - 5 years, 8 months ago

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Nice proof!

I didn't want to get too 'technical' for the CosinesGroup. I'm not too sure what they know, or are comfortable with.

Chung Kevin - 5 years, 8 months ago

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What's the Rational Root Theorem?

Pranav Kirsur - 5 years, 3 months ago

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It states that in a polynomial anxn+an1xn1++a1x+a0a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0, that all possible rational roots pq\dfrac{p}{q} have pa0p\mid a_0 and qanq\mid a_n.

Daniel Liu - 5 years, 1 month ago

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@Daniel Liu thanks

Pranav Kirsur - 5 years, 1 month ago

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What about proving the irrationality of ee? Obviously the above mentioned trick won't work. I've found a beautiful proof in "Proofs from THE BOOK", it's also featured on Wikipedia. Can anyone elaborate an alternative proof? What about irrationality of π\pi?

Nicolae Sapoval - 5 years, 8 months ago

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This article contains possibly the most elegant proof of the irrationality of pi that I've seen. You need only know integration by parts.

http://www.m-a.org.uk/resources/tmg/irrational_thoughts.doc

Enjoy.

David Treeby - 5 years, 8 months ago

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Thank you, it's indeed a really beautiful proof.

Nicolae Sapoval - 5 years, 8 months ago

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incase of root 3,let a=3m root 4 is rational

Aditya Khatavkar - 5 years, 8 months ago

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2.5 can be witiitn as 25/10 = 5/2.

0.33333… can be written as 3333.../10000…

mul and divide by 3, we get 9999.../(10000… x 3)

9999.. is close to 10000… and it is infinity so ratio is equal. = 1/3

if we were to write √2 as a ratio,

1414…/1000… and we won't know the last term or any properties of the huge number i.e divisible by 2, 3, or 5 something like that.

hence √2 cannot be expressed as a ratio

Soham Zemse - 5 years, 8 months ago

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That's an interesting observation, and is where I would like to go next.

The fact that you are trying to get to, is that a decimal number is rational if and only if it terminates, or is eventually repeating.

Chung Kevin - 5 years, 8 months ago

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by contradiction that it is rational

anurag mishra - 5 years, 8 months ago

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Can you elaborate? What do you mean?

Chung Kevin - 5 years, 8 months ago

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just take 2p=4q' and 4p'=2q and the divide the make 4 eq compare them u will get the answer

Adil Ahmed Khan Tanoli - 5 years, 8 months ago

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I don't think that we can prove that 4\sqrt{4} is irrational . . . For 3\sqrt{3}, I would think that this could be proven using a 30-60-90 triangle and the Pythagorean Theorem, but I don't have time (I have Fermat's Procrastination Disease). Anyone up to the challenge?

Josh Speckman - 5 years, 8 months ago

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Can you prove that 4 \sqrt{4} is rational then?

Hm, a geometric interpretation of a rational number is slightly harder to arrive at, as the concept of rational numbers is a number theoretic concept.

Chung Kevin - 5 years, 8 months ago

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by this method a/b is not a rational number since those two (a and b)are even
here i had a doubt
consider 6/4 where both 6 and 4 are even 6/4=3/2 which is a rational number thus my doubt is about the definition of a rational number any one can explain this?????????????

Lokesh Naani - 5 years, 8 months ago

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Note that rational square root of an integer has to be an integer in its reduced form. Because if N=pq\sqrt{N}=\frac{p}{q} with q>1q>1 and qpq\nmid p and gcd(p,q)=1gcd(p,q)=1, we have N=p2q2N= \frac{p^2}{q^2}. Hence q2p2q^2|p^2, which is impossible unless qpq|p. With the above in view, we just check that 12=11^2=1 and 22=42^2=4 and hence there is no integer (and hence no rational number) whose square root is 3.

Abhishek Sinha - 5 years, 8 months ago

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If this question is too easy for you, leave a hint instead of revealing the entire solution.

Chung Kevin - 5 years, 8 months ago

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