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Root 2 is irrational

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

A rational number is a number that can be written as $$\frac{a}{b}$$, where $$a$$ and $$b$$ 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 $$\sqrt{2}$$ is irrational:

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

Squaring the equation, we get that $$a^2 = 2 b^2$$. Since $$a^2$$ is even, hence $$a$$ is even. Let $$a = 2x$$, where $$x$$ is an integer. Substituting this in, we get $$4x^2 = 2b^2$$, or that $$2x^2 = b^2$$. We repeat the above argument. Since $$b^2$$ is even, thus $$b$$ is even.

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

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

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

Note by Chung Kevin
3 years, 3 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 $$p$$ is prime, then $$\sqrt{p}$$ is irrational.

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

Because 2 is prime, it follows that $$\sqrt{2}$$ is irrational. · 3 years, 3 months ago

It's much simpler to consider the polynomial $$x^2-2=0$$. 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 $$x=\sqrt{2}$$ · 3 years, 3 months ago

What's the Rational Root Theorem? · 2 years, 11 months ago

It states that in a polynomial $$a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0$$, that all possible rational roots $$\dfrac{p}{q}$$ have $$p\mid a_0$$ and $$q\mid a_n$$. · 2 years, 9 months ago

thanks · 2 years, 9 months ago

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. · 3 years, 3 months ago

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 · 3 years, 3 months ago

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. · 3 years, 3 months ago

incase of root 3,let a=3m root 4 is rational · 3 years, 3 months ago

What about proving the irrationality of $$e$$? 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 $$\pi$$? · 3 years, 3 months ago

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. · 3 years, 3 months ago

Thank you, it's indeed a really beautiful proof. · 3 years, 3 months ago

by this method a/b is not a rational number since those two (a and b)are even
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????????????? · 3 years, 3 months ago

I don't think that we can prove that $$\sqrt{4}$$ is irrational . . . For $$\sqrt{3}$$, 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? · 3 years, 3 months ago

Can you prove that $$\sqrt{4}$$ 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. · 3 years, 3 months ago

just take 2p=4q' and 4p'=2q and the divide the make 4 eq compare them u will get the answer · 3 years, 3 months ago

Comment deleted Dec 14, 2013

If this question is too easy for you, leave a hint instead of revealing the entire solution. · 3 years, 3 months ago

by contradiction that it is rational · 3 years, 3 months ago

Can you elaborate? What do you mean? · 3 years, 3 months ago

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