Prove 2^(1/2) + 6^(1/2) is irrational?

**mfetch22** · July 22nd 2010, 10:17 AM

I know how to prove that $\sqrt{2}$ and $\sqrt{6}$ are both themselves irrational, but how would you go about proving that

$\sqrt{2} + \sqrt{6}$ is irrational?

I've tried making the assumption that $\sqrt{2} + \sqrt{6} = \frac{a}{b}$ where $a$ and $b$ are both natural numbers, but it hasn't gotten me anywhere yet. I've manipulated the expression and made some observations when I see certain expressions in the form of $2n$ or $2n+1$ , therefore being either even or odd, but I can't seem to get to a proper proof. I have though, after making the rational assumption, run into multiple "contradictions", things that just can't be equal. But I'm not certain if these are the type of contradictions that count towards proving the irrationality, or if I made a simple mistake to arrive at the contradictions.

I'm also trying to prove the irrationality of $\sqrt{2} + \sqrt{3}$ , its next to the other question in my textbook, and I also have had the same experience with this number as well. thanks in advance for any help.

**melese** · July 22nd 2010, 11:02 AM

Here is a hint. Starting in similar way as you did, assume that $\sqrt{2} +\sqrt{6}=r$ , where $r$ is some rational number.
Now consider $r^2$ ...

**Unbeatable0** · July 22nd 2010, 02:44 PM

I found this question rather interesting, as I have never actually thought about what operations

we can apply on certain classes of irrational numbers (as in the class of all irrational

square roots of rational numbers), and still get a rational number.

Although I support melese's method, it will not always succeed in telling when $\sqrt{a}+\sqrt{b}$ is irrational.

For example when $a=3\ ,\ b=12$ (although it perfectly solves the original problem and I suggest

that you use that method in your excercise, since that's what was probably meant to be used).

Here's a nice way (in my opinion) to prove that if $\sqrt{a}\not\in\mathbb{Q}$

with $a,b\in\mathbb{Q}$ , then $\sqrt{a}+\sqrt{b}$ is irrational.

Suppose the contrary. Then $\sqrt{a}+\sqrt{b} \in \mathbb{Q}$ . Therefore

$\sqrt{a}-\sqrt{b} = \frac{a-b}{\sqrt{a}+\sqrt{b}} \in \mathbb{Q}$

But then $\sqrt{a} = \frac{1}{2}(\sqrt{a}-\sqrt{b})+\frac{1}{2}(\sqrt{a}+\sqrt{b}) \in \mathbb{Q}$ , a contradiction.

**roninpro** · July 22nd 2010, 10:00 PM

You can also note that $\sqrt{2}+\sqrt{6}$ is a root of the polynomial $p(x)=16-16 x^2+x^4$ . If $p$ has any rational roots, then they must be of the form $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$ . But we can see that none of these are roots of $p$ . Therefore, $\sqrt{2}+\sqrt{6}$ is irrational.

**Dinkydoe** · July 23rd 2010, 02:38 AM

You can also note that is a root of the polynomial . If has any rational roots, then they must be of the form . But we can see that none of these are roots of . Therefore, is irrational

Or in general: $\sqrt{a}+\sqrt{b}$ has minimal polynomial $p(x) =x^4-2(a+b)x^2+(a-b)^2$ (wich can be found by looking at powers of $\sqrt{a}+\sqrt{b}$ )

If $p(x)$ has any rational roots, it must be a divisor $d$ of $(a-b)^2$ . Now, we need to check that any divisor $d$ can not correspond with $\sqrt{a}+\sqrt{b}$ . Thus showing $\sqrt{a}+\sqrt{b}$ is not an integer is also enough.

Still, Unbeatable0's method is cutest

**mfetch22** · July 23rd 2010, 05:02 AM

Originally Posted by melese

Here is a hint. Starting in similar way as you did, assume that $\sqrt{2} +\sqrt{6}=r$ , where $r$ is some rational number.
Now consider $r^2$ ...

I've just spent some time trying to find the proof, with little success, after noting $\sqrt{2} + \sqrt{6} = \frac{a}{b} = r$ for some rational $r$ , and then considering $r^2$ as you said. I'm sure the proof is right under my nose, and I'm just not seeing it, do you have another hint? This is how far I got (granted, I could be "moving" in the wrong direction with these steps):

$\sqrt{2} + \sqrt{6} = \frac{a}{b} = r$ therefore

$r^2 = \frac{a^2}{b^2} = 8+2\sqrt{12} = 2(4 + \sqrt{12})$ that leads to this equation:

$(\frac{a^2}{2b^2} - 4)^2 = 12$

this led me to:

$(\frac{a^2}{2b^2})^2-8(\frac{a^2}{2b^2})+4 = 0$

But I feel like this is to complicated to be anywhere in the right direction, I did the isolation of the $\sqrt{12}$ and squared both sides in an attempt to get rid of irrationals so that I could use natural number factors to prove a common divisor between $a$ and $b$ .

**undefined** · July 23rd 2010, 05:19 AM

Originally Posted by mfetch22

I've just spent some time trying to find the proof, with little success, after noting $\sqrt{2} + \sqrt{6} = \frac{a}{b} = r$ for some rational $r$ , and then considering $r^2$ as you said. I'm sure the proof is right under my nose, and I'm just not seeing it, do you have another hint? This is how far I got (granted, I could be "moving" in the wrong direction with these steps):

$\sqrt{2} + \sqrt{6} = \frac{a}{b} = r$ therefore

$r^2 = \frac{a^2}{b^2} = 8+2\sqrt{12} = 2(4 + \sqrt{12})$ that leads to this equation:

$(\frac{a^2}{2b^2} - 4)^2 = 12$

this led me to:

$(\frac{a^2}{2b^2})^2-8(\frac{a^2}{2b^2})+4 = 0$

But I feel like this is to complicated to be anywhere in the right direction, I did the isolation of the $\sqrt{12}$ and squared both sides in an attempt to get rid of irrationals so that I could use natural number factors to prove a common divisor between $a$ and $b$ .

A series of subproofs by contradiction should suffice.

You got

$r^2 = 8+2\sqrt{12}$

Since by assumption r in Q, then r^2 in Q. Then 8 + 2*sqrt(12) is rational if and only if sqrt(12) is rational. (Prove by contradiction.) Then you can prove sqrt(12) is irrational in the manner that sqrt(2) is typically shown to be irrational. (Again, contradiction.) Alternatively, rewrite sqrt(12) as 2*sqrt(3) and prove sqrt(3) is irrational.

EDIT: melese's post #9 simplifies this by removing the first contradiction step.

**Bingk** · July 23rd 2010, 05:27 AM

Um, I think melese's method works for $\sqrt{3} + \sqrt{12}$

$\sqrt{3} + \sqrt{12} = r$ where $r$ is rational.

$r^2 = (\sqrt{3} + \sqrt{12})^2$
$r^2 = 3+12+2 \sqrt{36}$
$r^2 = 3+12+12$
$r^2 = 27$
$r = 3 \sqrt{3}$
$r_a = \sqrt{3}$ , where $r_a$ is rational

After that, it's just a matter of proving that $\sqrt{3}$ is irrational

Also, be careful when you say that $a,b\in\mathbb{Q}$ , because if $a$ and $b$ could be equal to $c^2$ and $d^2$ respectively, where $c,d\in\mathbb{Q}$ , respectively ... then $\sqrt{a} + \sqrt{b} = c + d$ .

**melese** · July 23rd 2010, 05:32 AM

Originally Posted by mfetch22

I've just spent some time trying to find the proof, with little success, after noting $\sqrt{2} + \sqrt{6} = \frac{a}{b} = r$ for some rational $r$ , and then considering $r^2$ as you said. I'm sure the proof is right under my nose, and I'm just not seeing it, do you have another hint? This is how far I got (granted, I could be "moving" in the wrong direction with these steps):

$\sqrt{2} + \sqrt{6} = \frac{a}{b} = r$ therefore

$r^2 = \frac{a^2}{b^2} = 8+2\sqrt{12} = 2(4 + \sqrt{12})$ that leads to this equation:

$(\frac{a^2}{2b^2} - 4)^2 = 12$

this led me to:

$(\frac{a^2}{2b^2})^2-8(\frac{a^2}{2b^2})+4 = 0$

But I feel like this is to complicated to be anywhere in the right direction, I did the isolation of the $\sqrt{12}$ and squared both sides in an attempt to get rid of irrationals so that I could use natural number factors to prove a common divisor between $a$ and $b$ .

To mfetch22, the direction I meant was very similar to what the user undefined did.
$r^2=8+\sqrt12;\ (r^2-8)/2=\sqrt12$ . Then clearly the left side is rational, but $\sqrt12$ is not.
An imprtant thing to notice is that I used $r$ instead of $a/b$ because the latter can distract the computation.

**Unbeatable0** · July 23rd 2010, 05:32 AM

Originally Posted by Bingk

Um, I think melese's method works for $\sqrt{3} + \sqrt{12}$

$\sqrt{3} + \sqrt{12} = r$ where $r$ is rational.

$r^2 = (\sqrt{3} + \sqrt{12})^2$
$r^2 = 3+12+2 \sqrt{36}$
$r^2 = 3+12+12$
$r^2 = 27$
$r = 3 \sqrt{3}$
$r_a = \sqrt{3}$ , where $r_a$ is rational

After that, it's the regular way of proving an irrational is not rational ...

Also, be careful when you say that $a,b\in\mathbb{Q}$ , because if $a$ and $b$ could be equal to $c^2$ and $d^2$ respectively, where $c,d\in\mathbb{Q}$ , respectively ... then $\sqrt{a} + \sqrt{b} = c + d$ .

That's right, but I considered the way I think melese meant it to be. i.e., getting to a contradiction $\sqrt{ab}\in\mathbb{Q}$ .

In my proof I assumed $\sqrt{a}\not\in\mathbb{Q}$ so it's not possible that $a=c^2\ ,\ c\in\mathbb{Q}$

**mfetch22** · July 23rd 2010, 05:54 AM

Thanks guys, I figured it out now, its so obvious now that I see the proof, but I guess thats how if often happens. Thanks again for everybodies help.

**Archie Meade** · July 23rd 2010, 06:42 AM

Originally Posted by mfetch22

I've just spent some time trying to find the proof, with little success, after noting $\sqrt{2} + \sqrt{6} = \frac{a}{b} = r$ for some rational $r$ , and then considering $r^2$ as you said. I'm sure the proof is right under my nose, and I'm just not seeing it, do you have another hint? This is how far I got (granted, I could be "moving" in the wrong direction with these steps):

$\sqrt{2} + \sqrt{6} = \frac{a}{b} = r$ therefore

$r^2 = \frac{a^2}{b^2} = 8+2\sqrt{12} = 2(4 + \sqrt{12})$ that leads to this equation:

$(\frac{a^2}{2b^2} - 4)^2 = 12$

this led me to:

$(\frac{a^2}{2b^2})^2-8(\frac{a^2}{2b^2})+4 = 0$

But I feel like this is to complicated to be anywhere in the right direction, I did the isolation of the $\sqrt{12}$ and squared both sides in an attempt to get rid of irrationals so that I could use natural number factors to prove a common divisor between $a$ and $b$ .

Why not express $\sqrt{6}{\sqrt{2}=\sqrt{2}\sqrt{3}\sqrt{2}=2\sqrt{ 3}$

Now you only need prove $\sqrt{3}$ is irrational.

$\sqrt{3}=\frac{x}{y}\ ?$

$3=\frac{x^2}{y^2}\ \Rightarrow\ x^2=3y^2$

If $x^2$ is divisible by 3, then so is x, since the first square divisible by 3 is 9.

You can also show that y is divisible by 3, hence $\sqrt{3}$

cannot be expressed as an irreducible fraction.

It can be expressed as $\sqrt{3}=\frac{3\sqrt{3}}{3}=\frac{9\sqrt{3}}{9}\ etc$

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