# Irrational Numbers

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 cancelled 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 cancelled 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?

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#### dextercioby

Homework Helper
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 cancelled 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.

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

Homework Helper
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.

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

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#### dextercioby

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

Daniel.

#### Zurtex

Homework Helper
dextercioby said:
Because $p^{2}$ divisible by 3 DOES NOT IMPLY $p$ divisible by 3...

Daniel.
If 3 is a prime it does.

#### vincentchan

dextercioby said:
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.... ....

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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

dextercioby said:
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

Staff Emeritus
Gold Member
?

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

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 &radic;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.

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 &radic;2 + &radic;3, so it's irrational.

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

Staff Emeritus
Gold Member
But why does it imply it?

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

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#### dextercioby

Homework Helper
vincentchan said:
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...

Thanks guys (thanks Hurkyl for your wonderful explanation)

Ok so lets 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

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#### Justin Lazear

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

Daniel.

P.S.So i was right...
p was assumed to be an integer.

--J

#### Hurkyl

Staff Emeritus
Gold Member
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

Homework Helper
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.

#### Curious3141

Homework Helper
dextercioby said:
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.

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

Staff Emeritus
Gold Member
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

Homework Helper
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

dextercioby said:
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.

--J

#### Hurkyl

Staff Emeritus
Gold Member
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 &radic;2 + &radic;3. But, by taking other roots, we get three conjugates:

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

Homework Helper
dextercioby said:
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.

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

Homework Helper
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.

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