Prove that sqrt2 + sqrt6 is irrational.
Where do I start with this?
Or alternatively, consider the integer polynomial that has that number as one of its roots.
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
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.
Solve for sqrt(3)!
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.
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.
So that's not right...
Clearly, you want me to solve the polynomial you wrote. But then I get,
What nonsense is this!
Am I getting anywhere near a proof of anything besides my ignorance?
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
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
Does that suffice?
How can I know that the additive inverse of an irrational is also irrational?
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.
As for the integers being closed under + and *, you should be able to assume that to be true without proof.
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?
If I had to guess, I think they mean:
41 + Σ2n
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