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put s={f(pi^5) such that f(x) is an element of Q[x]}

Scary

solved this now

(edited 7 years ago)

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Reply 1

ghostwalker

Original post by Scary

Put s={

f(\pi^5):f(x) \in \mathbb{Q(X)}

} prove that the set s is not a subfield of

\mathbb{C}

Not an expert on this, so anyone more qualified feel free....

The a_0, constant, term isn't multiplied by anything.

f defined by f(x) = 1 meets the criterion for the set.

And so,

f(\pi^5)=1

and

1\in S

And yes you only need one axiom to fail.

(edited 7 years ago)

Reply 2

Scary

Original post by ghostwalker

Not an expert on this, so anyone more qualified feel free....

The a_0, constant, term isn't multiplied by anything.

f defined by f(x) = 1 meets the criterion for the set.

And so,

f(\pi^5)=1

and

1\in S

And yes you only need one axiom to fail.

ahh yeah sorry i was getting confused with the way in which the set was written so thanks for this, would the best approach be to write a general polynomial for f(pi^5) like when proving that it is transcendent only though in this case it would not be evaluated to give a finite value, and then test the axioms on this polynomial?

Reply 3

ghostwalker

Original post by Scary

Well you'd need more than one polynomial for most of the axioms, but I wouldn't go to that level of detail.

E.g
Additive closure, If

a,b\in S

, then

\exists f,g\in\mathbb{Q}(X)

such that

a=f(\pi^5),b=g(\pi^5)

Then

a+b = f(\pi^5)+g(\pi^5)=(f+g)(\pi^5)

where f+g is the standard sum formed by adding corresponding coefficients, and so

f+g\in\mathbb{Q}(X)

, hence

a+b\in S

Best I can see, there is only one axiom the set fails on.

(edited 7 years ago)

Reply 4

Scary

Original post by ghostwalker

Well you'd need more than one polynomial for most of the axioms, but I wouldn't go to that level of detail.

E.g
Additive closure, If

a,b\in S

, then

\exists f,g\in\mathbb{Q}(X)

such that

a=f(\pi^5),b=g(\pi^5)

Then

a+b = f(\pi^5)+g(\pi^5)=(f+g)(\pi^5)

where f+g is the standard sum formed by adding corresponding coefficients, and so

f+g\in\mathbb{Q}(X)

, hence

a+b\in S

Best I can see, there is only one axiom the set fails on.

Thanks for this example, would the axiom that fails be the multiplicative inverse?

Reply 5

ghostwalker

Original post by Scary

Thanks for this example, would the axiom that fails be the multiplicative inverse?

It would.

I presume you've eliminated the others. If not, you may find it a useful exercise to confirm they hold.

Reply 6

Scary

Original post by ghostwalker

It would.

I presume you've eliminated the others. If not, you may find it a useful exercise to confirm they hold.

yeah, i've just shown that s satisfies the first 4 so i assumed that this must be the one that fails, although because of the notation i dont really know on how to start it. would we just consider on finding a 1/f(pi^5) and a contradiction would come somewhere in the working for it?

Reply 7

ghostwalker

Original post by Scary

would we just consider on finding a 1/f(pi^5)

I don't know what you mean by that.

In general:

Consider: What is the axiom in question - mathematically, in full?

Then what would you need to do to show it's false?

Now apply that to this situation.

Last post for today.

Reply 8

Scary

okay thanks for the help i'm slowly getting there

Reply 9

ghostwalker

Original post by Scary

okay thanks for the help i'm slowly getting there

No problem. Since you've been active here and not posted further, I presume you've now solved this.

Reply 10

Scary

Original post by ghostwalker

No problem. Since you've been active here and not posted further, I presume you've now solved this.

i haven't been able to solve it, I'm having some problems when trying to compute the inverse to show a contradiction, is it possible you could guide me a bit further? it seems more challenging then i first assumed, i have no problem proving a set is not a subfield of c its just the way that this set is i haven't done an example like this so it's quite confusing. Thanks again

Reply 11

ghostwalker

Original post by Scary

Lets see some detailed working then of what you have done and how far you've got.

Reply 12

Scary

Original post by ghostwalker

Lets see some detailed working then of what you have done and how far you've got.

Yeah of course here's my working when testing the axiom I am unsure if we need to show that there exists no inverse function or there does not exist 1/f(pi^5) sorry for the picture quality my phone isn't the greatest

(edited 7 years ago)

image.jpg460.4KB

Reply 13

ghostwalker

Original post by Scary

Neither.

And any

f\in\mathbb{Q}(X)

is bijective here, since the domain is the point

\{\pi^5\}

and its codomain is

\{f(\pi^5)\}

. But that's not relevant.

Look back at my post from last night. What are you trying to show in S? Forget about where the elements of S come from for now.

(edited 7 years ago)

Reply 14

DFranklin

Original post by Scary

Wrote a long enough post without realising ghostwalker had repleid that I don't want to discard it, but I'm going to spoiler it and suggest you first try to follow his lead.

Spoiler

Reply 15

Scary

Original post by ghostwalker

Neither.

And any

f\in\mathbb{Q}(X)

is bijective here, since the domain is the point

\{\pi^5\}

and its codomain is

\{f(\pi^5)\}

. But that's not relevant.

Look back at my post from last night. What are you trying to show in S? Forget about where the elements of S come from for now.

i would need to show that an element of s has no inverse which would mean that axiom 5 would fail as just one case is not true. say we have an element a which is in s then the inverse say b means that a*b=1 , is the condition for F(x) in Q[x] effecting the values we can take for x? or can we choose an element in s that is not in Q, if we proceed with a and b being the inverse of a we would have ab=1=f(pi^5) but you shown 1 is in the set s which is confusing , it's really confusing me but i want to get to grips so i can practice more with different sets similar to this

(edited 7 years ago)

Reply 16

Scary

Original post by DFranklin

Wrote a long enough post without realising ghostwalker had repleid that I don't want to discard it, but I'm going to spoiler it and suggest you first try to follow his lead.

Spoiler

Thanks for this response, in regards to the spoiler, part (a) if a has an inverse b then ab=1 but we have shown 1 is in s( the first axiom) Part (b) for alpha to be in s we have alpha=f(pi^5) but we cant choose f=0 as it is in Q and has no inverse so i dont see how you can use the first to parts to show a contradiction to (c) although we could choose alpha to be pi^5 then beta would be 1/pi^5 and 1/pi^5 is not in Q? so 1/pi^5=f(pi^5) maybe?

(edited 7 years ago)

Reply 17

DFranklin

Original post by Scary

Here are the pertinent facts:

If a has an inverse b then ab = 1.
If a, b are in S, then we can find polys p, q with p(pi^5) = a, q(pi^5) = b.

At this point I will merely observe that p(X)q(X) is a polynomial with some quite interesting properties when you plug in pi^5...