Prove that sqrt2 + sqrt6 is irrational

  • Thread starter StephenPrivitera
  • Start date
  • Tags
    Irrational
In summary: So, for example, if k=5:41 + 2 + 4 + 6 + 8 + 10 = 71In summary, to prove that sqrt2 + sqrt6 is irrational, we can assume otherwise and use the polynomial with that number as one of its roots. We can also use the theorem that if p/q is a rational root of an integer polynomial in lowest terms, then p divides the constant term and q divides the leading coefficient. Additionally, we can assume that the integers are closed under addition and multiplication. Another problem to consider is whether the sum 41 + Σ2n, from n=1 to n=k, is always prime.
  • #1
StephenPrivitera
363
0
Prove that sqrt2 + sqrt6 is irrational.
Where do I start with this?
 
Last edited by a moderator:
Mathematics news on Phys.org
  • #2
Assume otherwise!

Or alternatively, consider the integer polynomial that has that number as one of its roots.
 
  • #3
I thought about the first one. It doesn't suit me well.
Suppose it is rational. Then,
sqrt2 +sqrt6 = p/q
With some other rational-irrational proofs, I squared both sides. So,
2 +2sqrt12 +6 =p2/q2
4(sqrt3+4)=
(2q)2(sqrt3+4)=p2
Now what?

As for the second part, I'm not sure I know what you mean by "integer" polynomial. Do you mean a polynomial whose domain is the integers?

(x - (sqrt2 + sqrt6))(x - r)=0
x^2 - x(sqrt2 +sqrt6) -xr +r(sqrt2 + sqrt6) = 0
(sqrt2 + sqrt6)(x -r) = x(x-r)
Let's hope x does not = r.
Then sqrt2 + sqrt6 = x
...
This of course makes perfect sense, since x doesn't equal r, it must equal it's other root. What pointless work! Certainly, I don't understand what you mean.
 
  • #4
(2q)2(sqrt3+4)=p2

Now what?

Solve for sqrt(3)!




As for the second part, I'm not sure I know what you mean by "integer" polynomial. Do you mean a polynomial whose domain is the integers?

I meant one whose coefficients are integers...

x = sqrt(2) + sqrt(6)
x^2 = 8 + 4 sqrt(3)
x^2 - 8 = 4 sqrt(3)
x^4 - 16 x^2 + 64 = 48
x^4 - 16 x^2 + 16 = 0

And depending on how much you remember about solving polynomials, the result is trivial from here.
 
  • #5
Ok, wow, you're going to think I'm really incompetent. I really don't understand what to do for either instance. Sure, I can solve for sqrt3, but then what?
You can square both sides and you get something quite similar to your second method.
3=(p/2q)^4 - 8(p/2q)^2 +16
or 0=(p/2q)^4 - 8(p/2q)^2 +13
I could solve this, but the result is obvious. I'd just be undoing the square I just applied to both sides.
(p/2q)^2=4+sqrt3
So that's not right...

Clearly, you want me to solve the polynomial you wrote. But then I get,
x2=(16+8sqrt3)/2=8+4sqrt3=4(2+sqrt3)
So, x=2sqrt(2+sqrt3)
What nonsense is this!

Am I getting anywhere near a proof of anything besides my ignorance?
Please help.
 
  • #6
Here's another one I can't get.
If a is rational and b is irrational is a+b necessarily irrational? I say yes, but suppose not. Then,
a+b=p/q for some p and q in J
and a=c/d for some c,d in J
then b=p/q-c/d=(pd-cq)/(pd)
Now if we can believe that the integers are closed under * and +, then we have derived a contradiction since pd-cq will be an integer and pd will be an integer. But how can I show they in fact are closed under * and +?

What if a and b are both irrational?
Not necessarily because
pi+(-pi)=0
Does that suffice?
How can I know that the additive inverse of an irrational is also irrational?
 
  • #7
Ok, wow, you're going to think I'm really incompetent.

Nah, this is one of those things that you usually have to see a few times before the method sinks in.


When you solved for sqrt(3), you got:

sqrt(3) = (p/(2q))^2 - 4

We assumed that sqrt(2) + sqrt(6) is rational, and from that assumption we've proven that there is some integers p and q so the above equation holds... the left hand side of this equation is an irrational number, but what about the right hand side?


As for the integer polynomial, there is a theorem that if p/q is a rational root of an integer polynomial in lowest terms, then p divides the constant term and q divides the leading coefficient. For example, if the equation

2x^3 - 3x + 6 = 0

has any rational roots, they must be among these possibilities:

1/1, -1/1, 2/1, -2/1, 3/1, -3/1, 6/1, -6/1, 1/2, -1/2, 3/2, -3/2

so you just have to exhaust all the possibilities to prove that this polynomial has no rational roots.

So try this theorem with the polynomial

x^4 - 16 x^2 + 16 = 0

for which we know sqrt(2) + sqrt(6) is a root.
 
  • #8
As for the integers being closed under + and *, you should be able to assume that to be true without proof.
 
  • #9
Guess what? Problem number 18 in my text asks me to prove that very theorem about integer polynomials. It wasn't assigned so I might not get around to it. Sounds tough.

But how about this one:
prove or disprove: If n is a natural number, then
41 + 2 + 4 + 6 + ... + 2n is prime.

My main problem is understanding what the pattern is. To me this is like asking me to prove that any sum of numbers is prime. Can you see a pattern? Maybe the first term is supposed to be one (not 41)? Then the pattern would be 1+2+4+6+8+10+12+...+2n, right?
 
  • #10
If I had to guess, I think they mean:

41 + Σ2n
 

FAQ: Prove that sqrt2 + sqrt6 is irrational

1. What does it mean for a number to be irrational?

A number is considered irrational if it cannot be expressed as a ratio of two integers. In other words, it cannot be written as a fraction.

2. How do you prove that sqrt2 + sqrt6 is irrational?

To prove that sqrt2 + sqrt6 is irrational, we can use a proof by contradiction. Assume that sqrt2 + sqrt6 is rational and can be written as a fraction a/b, where a and b are integers with no common factors. Then we can square both sides of the equation to get 2 + 2sqrt3 + 6 = a^2/b^2. Simplifying, we get 8 + 2sqrt3 = a^2/b^2. This means that 2sqrt3 is also rational, which is a contradiction since sqrt3 is known to be irrational. Therefore, our initial assumption is false and sqrt2 + sqrt6 must be irrational.

3. Can you use a different method to prove that sqrt2 + sqrt6 is irrational?

Yes, there are multiple methods that can be used to prove that sqrt2 + sqrt6 is irrational. One common method is to use the rational root theorem, which states that if a polynomial with integer coefficients has a rational root, that root must be a factor of the constant term. We can apply this theorem to the polynomial x^2 - 8x + 8, which has sqrt2 + sqrt6 as a root. Since the only possible rational factors of 8 are 1, 2, 4, and 8, and none of these are roots of the polynomial, we can conclude that sqrt2 + sqrt6 is irrational.

4. Why is proving that sqrt2 + sqrt6 is irrational important?

Proving the irrationality of sqrt2 + sqrt6 is important because it is a fundamental concept in mathematics. It helps us understand the properties of irrational numbers and their relationships with rational numbers. This proof also serves as an example of how to use various mathematical techniques, such as proof by contradiction and the rational root theorem, to solve problems.

5. Are there any practical applications of this proof?

While the proof itself may not have any direct practical applications, the concept of irrational numbers and their properties are used in many fields, including physics, engineering, and computer science. For example, irrational numbers are often used in calculations involving measurements, such as in the design of buildings and bridges. Additionally, the proof of irrationality for sqrt2 + sqrt6 can be extended to other numbers, leading to a better understanding of the irrational number system as a whole.

Similar threads

MHB Finding the Distance from Point P to AC on an ABCD.EFGH Cube
Replies
4
Views
1K
MHB Calculating G's Distance from BH on an ABCD.EFGH Cube
Replies
2
Views
1K
B Proving the area formula for a rectangle for all positive real numbers
Replies
10
Views
543
MHB Simplifying radicals - Help with basic number manipulation
Replies
5
Views
3K
I Is the Ratio of Circumference to Radius of a Circle Always Irrational?
Replies
19
Views
2K
B Square Root of an Odd Powered Integer is Always Irrational?
Replies
31
Views
3K
I Is the odd root of an even number always an irrational number?
Replies
9
Views
1K
MHB What is the Value of \(2\tan\frac{1}{2}A\tan\frac{1}{2}B\) in Triangle ABC?
Replies
2
Views
934
MHB [ASK] Exponents and Roots Simplification problem
Replies
12
Views
2K
MHB Challenge involving irrational number
Replies
3
Views
1K

Hot Threads

Recent Insights

Back
Top