Proving the Irrationality of √3 and Other Non-Perfect Square Roots

[courtrigrad]

Hello all

I encountered a few questions on irrational numbers.

1. Prove that $$\sqrt{3}$$ is irrational [/tex]. So let $$l = \sqrt{3}$$. Then if $$l$$ were a rational number and equal to $$\frac{p}{q}$$ where $$p, q$$ are integers different from zero then we have $$p^{2} = 3q^{2}$$. We can assume that $$p, q$$ have no common factors, because they would be canceled out in the beginning. Now $$p^2$$ is divisible by 3. So let $$p = 3p'$$. We have $$9p'^2 = 3q^2$$ or $$q^2 = 3p'^{2}$$. So both $$p , q$$ are divisible by 3. But this contradicts the fact that common factors of $$p, q$$ were canceled out. Hence $$\sqrt{3}$$ is irrational.

2. If we had to prove that $$\sqrt{n}$$ was an irrational number where $$n$$ is not a perfect square would be do basically the same thing as we did above?

 

[dextercioby]

Hello all

I encountered a few questions on irrational numbers.

1. Prove that $$\sqrt{3}$$ is irrational [/tex]. So let $$l = \sqrt{3}$$. Then if $$l$$ were a rational number and equal to $$\frac{p}{q}$$ where $$p, q$$ are integers different from zero then we have $$p^{2} = 3q^{2}$$. We can assume that $$p, q$$ have no common factors, because they would be canceled out in the beginning. Now $$p^2$$ is divisible by 3. So let $$p = 3p'$$.

That's WRONG...And the rest of it is wrong as well...

2. If we had to prove that $$\sqrt{n}$$ was an irrational number where $$n$$ is not a perfect square would be do basically the same thing as we did above?

NO,the proof following your pattern works only for $\sqrt{2}$

Daniel.

 

[courtrigrad]

Ok so how would you do it for $$\sqrt{3}$$ or for any matter $$\sqrt{n}$$? What about if you had something like $$\sqrt{2} + \sqrt{3}$$

Thanks

 

[dextercioby]

Ok so how would you do it for $$\sqrt{3}$$ or for any matter $$\sqrt{n}$$? What about if you had something like $$\sqrt{2} + \sqrt{3}$$

Thanks

I have no idea...I'm not a mathematician.The error i spotted was just common sense...

Daniel.

 

[courtrigrad]

Why is what I have wrong? I just used proof by contradiction, as in $$\sqrt{2}$$

Would it be that $$p^2$$ has even powers of primes as its factors?

Thanks

 

[dextercioby]

Because $p^{2}$ divisible by 3 DOES NOT IMPLY $p$ divisible by 3...

Daniel.

 

[Zurtex]

Because $p^{2}$ divisible by 3 DOES NOT IMPLY $p$ divisible by 3...

Daniel.

If 3 is a prime it does.

 

[vincentchan]

Because $p^{2}$ divisible by 3 DOES NOT IMPLY $p$ divisible by 3...

Daniel.

dexterciogy, u were wrong
can you give me a conter-example, then

ps. i didn't use large font this time... ...

 

[courtrigrad]

Ok let me take a shot at this again

$$\sqrt{3} = \frac{p}{q}$$ $$3 = \frac{p^2}{q^2}$$ So $$p^2 = 3q^2$$. So this this mean that $$p^2$$ has factors of even powers of primes hence $$3^2$$ is a factor which implies $$3$$ is a factor. Same with $$q^2$$

This is a guess

Thanks

 

[Justin Lazear]

That's WRONG...

I understand that I generally lack common sense, but what's WRONG with it?

For the sum of the two roots, square the sum first.

--J

 

[Hurkyl]

?

This method of proof works fine for any prime...

But I want to ask courtrigrad:

Why does p^2 = 3q^2 let you conclude that p is divisible by 3?



For any irrational algebraic number, this method should allow you to prove it's irrational:

If a is an algebraic number, then let f(x) be the minimum polynomial of a. That is, the smallest polynomial such that f(a)=0. (Actually, you don't need the smallest, there is a wide class of polynomials that would work)

Now, you should know the criterion for a rational number to be a root of a polynomial: the only candidates are those whose numerator divides the constant term, and whose denominator divides the leading term.

The minimum polynomial of √3 is x^2 - 3 = 0. The only rational numbers that could be a root are 1, -1, 3, -3. Obviously, none of them work.

for √2 + √3, let it equal a...
a^2 = 5 + 2√6
a^2 - 5 = 2√6
a^4 - 10a^2 + 25 = 24
a^4 - 10a^2 + 1 = 0

The minimum polynomial of a is x^4 - 10x^2 + 1. (I haven't proven it actually is the minimum, but it will still suffice for this method of proof)

Now, the only possible rational roots of this are 1 and -1, and neither of these is √2 + √3, so it's irrational.

 

[courtrigrad]

because $$p^2$$ has $$3^2$$ as one of its factors which implies that $$3$$ is a factor of $$p$$. Is this right?

Thanks

 

[Hurkyl]

But why does it imply it?

Recall this theorem: if p is prime, and p | ab, then p | a or p | b

 

[dextercioby]

dexterciogy, u were wrong
can you give me a conter-example, then

ps. i didn't use large font this time... ...

$$p^{2}=3\cdot 19 \Rightarrow p=\sqrt{57}$$ which is not divisible by 3...

Daniel.

P.S.So i was right...

 

[courtrigrad]

Thanks guys (thanks Hurkyl for your wonderful explanation)

Ok so let's say I have $$\sqrt{2} + ^3\sqrt{2}$$ and we want to prove that it's irrational. Dp I just raise this to the sixth power and work off from here?

Also if we have $$^3\sqrt{3}$$ and we want to prove that its irrational I receive $$p^3 = 3q^3$$. Would I emply the same reasoning as the other problems? Would it still be factors of even powers of primes ?

Thanks

 

[Justin Lazear]

$$p^{2}=3\cdot 19 \Rightarrow p=\sqrt{57}$$ which is not divisible by 3...

Daniel.

P.S.So i was right...

p was assumed to be an integer.

--J

 

[Hurkyl]

Daniel: remember that p was constructed to be an integer.


courtrigrad: the usual trick is to take the first few powers of the number, compute a few of its powers, and try to add them to get zero, and use that to get the minimum polynomial.

I'd expect that you'd need to compute 6 powers of your number. (because it's the sum of a square root and a cube root) So, if a is your number, compute 1, a, a^2, a^3, a^4, a^5, and a^6, and try to find a linear combination of them (i.e. you can multiply by constants and add them... much like elementary row operations on a matrix) that equals zero. Then, you have a polynomial with a as a root, and you can find all possible rational numbers that could be a rood.

 

[dextercioby]

I knew that.The idea was that everybody (vincentchan,zurtex and justin) argued that my assertion

p^{2}=3k =/=> p=3k'

was wrong...

I showed them i was right...

Daniel.

P.S.This thread cost me another black ball with the leaders...

 

[Curious3141]

I knew that.The idea was that everybody (vincentchan,zurtex and justin) argued that my assertion

p^{2}=3k =/=> p=3k'

was wrong...

I showed them i was right...

Daniel.

P.S.This thread cost me another black ball with the leaders...

For the love of Zarathustra,

p has always been assumed to be an integer ! Your example with $\sqrt{57}$ proves nothing ! Yet you smugly assert you were right all along. :yuck:

Do you understand the concept of a proof by contradiction ? The only problem that I could find with the OP's proof was that he didn't assert that p and q were coprime integers to begin with. Other than that, the proof stands, as it would for any prime n.

Your objection to his proof was unfounded and wrong all along.

 

[Hurkyl]

Here, let me redo my example the cookie-cutter way.

$$\alpha := \sqrt{2} + \sqrt{3}$$
$$\alpha^2 = 5 + 2\sqrt{6}$$
$$\alpha^3 = 11 \sqrt{2} + 9 \sqrt{3}$$
$$\alpha^4 = 49 + 20 \sqrt{6}$$

I want to find an (integer) linear combination of the numbers $1, \alpha, \alpha^2, \alpha^3, \alpha^4$ that equals zero.

So, I apply linear algebra. You can consider a module over the integers (a generalization of a vector space) and apply linear algebra. The "basis vectors" are $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$, so I want to solve the system:

$$\left( \begin{array}{ccccc} 1 & 0 & 5 & 0 & 49 \\ 0 & 1 & 0 & 11 & 0 \\ 0 & 1 & 0 & 9 & 0 \\ 0 & 0 & 2 & 0 & 20 \end{array} \right) \vec{x} = \left( \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array} \right)$$

Note that we have 5 unknowns and 4 equations, so the system is underdetermined and must have a nontrival solution.

Of course, you could do this by inspection too. I would start with $\alpha^4 - 10 \alpha^2$ which reduces to -1, so we have the combination: $\alpha^4 - 10\alpha^2 + 1 = 0$

So, this proves $\alpha$ is a root of the polynomial $x^4 - 10x^2 + 1$. Then, you simply exhaust over the candidates for a rational root of this polynomial, and show that none exist. Thus, $\alpha$ is irrational.

 

[dextercioby]

Yes,it may have been.The key point is that the three posters which contradicted me didn't do it on the error itself,but on something else which was incidentally true...

Daniel.

 

[Justin Lazear]

Yes,it may have been.The key point is that the three posters which contradicted me didn't do it on the error itself,but on something else which was incidentally true...

Daniel.

*scratches head*

--J

 

[Hurkyl]

Another way of finding this polynomial works like this:

You know that generally, when you take a square root, you get two different values... well, we can look at the "conjugates" of our number.

Our number was √2 + √3. But, by taking other roots, we get three conjugates:

-√2 + √3
√2 - √3
-√2 - √3

It turns out that the minimum polynomial has, as its roots, precisely this group of conjugate numbers. Thus, the polynomial can be factored over the reals as:

$$(x - \sqrt{2} - \sqrt{3})(x - \sqrt{2} + \sqrt{3})(x + \sqrt{2} - \sqrt{3})(x + \sqrt{2} + \sqrt{3})$$

When you expand it, you get (yet again) the polynomial $x^4 - 10x^2 + 1$.


The number you have has 5 conjugates: two choices for the square root, and 3 choices for the cube root yields 6 numbers in all. You could find the minimum polynomial of your number as I did in this post.

 

[Zurtex]

I knew that.The idea was that everybody (vincentchan,zurtex and justin) argued that my assertion

p^{2}=3k =/=> p=3k'

was wrong...

I showed them i was right...

Daniel.

P.S.This thread cost me another black ball with the leaders...

Actually I'm going to stand by my point here, you say:

$$p^{2}=3\cdot 19 \Rightarrow p=\sqrt{57}$$


However ignoring the obvious fact in the first place that we were assuming p to be in an integer and that if p² is divisable by 3 then p is divisable by 3 for all p in Z, you are clearly not talking about integers so the square root of 57 is divisable by 3 as:

$$3 \cdot \sqrt{\frac{57}{9}} = p$$

 

[HallsofIvy]

This has gone on for a while, but:

Any integer, p, must be of one of these forms: 3n, 3n+1, 3n+2 for some integer n.

If p= 3n+1 then p²= 9n²+ 6n+ 1= 3(3n²+ 2n)+ 1.

If p= 3n+2 then p²= 9n²+ 12n+ 4= 9n²+ 12n+ 3+1
= 3(3n²+4n+ 1)+ 1

If p= 3n then p²= 9n²= 3(3n²)

That is, p² is a multiple of 3 only if p itself is a multiple of 3:
For any integer, p² divisible by 3 implies p = 3k for some integer k.

 

[Galileo]

It's senseless to talk about divisibility if you take real numbers. Any real number is divisible by any other real number except zero, in the sense that the quotient is also a real number.