# Homework Help: Irrational Numbers

1. Feb 1, 2005

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?

2. Feb 1, 2005

### dextercioby

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

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

Daniel.

3. Feb 1, 2005

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

4. Feb 1, 2005

### dextercioby

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

Daniel.

5. Feb 1, 2005

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

Last edited: Feb 1, 2005
6. Feb 1, 2005

### dextercioby

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

Daniel.

7. Feb 1, 2005

### Zurtex

If 3 is a prime it does.

8. Feb 1, 2005

### vincentchan

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

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

Last edited: Feb 1, 2005
9. Feb 1, 2005

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

10. Feb 1, 2005

### Justin Lazear

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

11. Feb 1, 2005

### Hurkyl

Staff Emeritus
?

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.

12. Feb 1, 2005

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

Thanks

13. Feb 1, 2005

### Hurkyl

Staff Emeritus
But why does it imply it?

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

Last edited: Feb 1, 2005
14. Feb 1, 2005

### dextercioby

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

Daniel.

P.S.So i was right...

15. Feb 1, 2005

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

Last edited: Feb 1, 2005
16. Feb 1, 2005

### Justin Lazear

p was assumed to be an integer.

--J

17. Feb 1, 2005

### Hurkyl

Staff Emeritus
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.

18. Feb 1, 2005

### 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.

19. Feb 1, 2005

### Curious3141

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.

20. Feb 1, 2005

### Hurkyl

Staff Emeritus
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.