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

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Scary
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solved this now
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ghostwalker
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(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.
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(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?
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(Original post by Scary)
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?
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.
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(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?
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(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.
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(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?
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(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.
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okay thanks for the help i'm slowly getting there
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(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.
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(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
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(Original post by Scary)
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
Lets see some detailed working then of what you have done and how far you've got.
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(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
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(Original post by Scary)

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
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.
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(Original post by Scary)
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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
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:
Show

Sorry, but this all looks a bit confused.

A few points:

You're wanting to disprove that every non-zero element has an inverse. To do that, you just need to find one element that doesn't have an inverse. In this case, anything element of S that isn't in Q is going to work. I think you would find it easier if you picked one element (not in Q) and tried to show it has no inverse. There's a fairly obvious "simple" choice to make, but to leave things open I'll just call the element \alpha

Then you really just need to concentrate on 3 things:

(a) What does it mean for \beta to be an inverse of \alpha?
(b) What does it mean for \alpha or \beta to be in S?
(c) Since \pi^5 is transcendental, is it possible to have a non-zero polynomial p(x) with p(\pi ^5) = 0?

You can combine (a) and (b) to get a contradiction to (c).
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(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
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(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:
Show



Sorry, but this all looks a bit confused.

A few points:

You're wanting to disprove that every non-zero element has an inverse. To do that, you just need to find one element that doesn't have an inverse. In this case, anything element of S that isn't in Q is going to work. I think you would find it easier if you picked one element (not in Q) and tried to show it has no inverse. There's a fairly obvious "simple" choice to make, but to leave things open I'll just call the element \alpha

Then you really just need to concentrate on 3 things:

(a) What does it mean for \beta to be an inverse of \alpha?
(b) What does it mean for \alpha or \beta to be in S?
(c) Since \pi^5 is transcendental, is it possible to have a non-zero polynomial p(x) with p(\pi ^5) = 0?

You can combine (a) and (b) to get a contradiction to (c).


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?
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(Original post by Scary)
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?
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...
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(Original post by DFranklin)
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...
I understand this now thank you, is the contradiction that p(pi^5)q(pi^5)=1 but we know that p(pi^5) is transcendental so it cannot equal one?
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(Original post by Scary)
I understand this now thank you, is the contradiction that p(pi^5)q(pi^5)=1 but we know that p(pi^5) is transcendental so it cannot equal one?
No, that's not right. That equation doesn't imply that p(pi^5) = 1, also,how do you know that p(pi^5) is transcendental?
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