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

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#2

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

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

And so, and

And yes you only need one axiom to fail.

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(Original post by

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, and

And yes you only need one axiom to fail.

**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, and

And yes you only need one axiom to fail.

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#4

(Original post by

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?

**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?

E.g

Additive closure, If , then such that

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

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

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(Original post by

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 , then such that

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

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

**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 , then such that

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

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

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#6

(Original post by

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

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

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

It would.

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

**ghostwalker**)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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#8

(Original post by

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

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

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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#10

(Original post by

okay thanks for the help i'm slowly getting there

**Scary**)okay thanks for the help i'm slowly getting there

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(Original post by

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

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

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#12

(Original post by

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

**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

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(Original post by

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

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

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#14

(Original post by

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

**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

And any is bijective here, since the domain is the point and its codomain is . 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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#15

**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

Spoiler:

Sorry, but this all looks a bit confused.

A few points:

You're wanting to

Then you really just need to concentrate on 3 things:

(a) What does it

(b) What does it mean for or to be in S?

(c) Since is transcendental, is it possible to have a non-zero polynomial p(x) with ?

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

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

Then you really just need to concentrate on 3 things:

(a) What does it

**mean**for to be an inverse of ?

(b) What does it mean for or to be in S?

(c) Since is transcendental, is it possible to have a non-zero polynomial p(x) with ?

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

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(Original post by

Neither.

And any is bijective here, since the domain is the point and its codomain is . 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.

**ghostwalker**)Neither.

And any is bijective here, since the domain is the point and its codomain is . 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

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.

**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:

Sorry, but this all looks a bit confused.

A few points:

You're wanting to

Then you really just need to concentrate on 3 things:

(a) What does it

(b) What does it mean for or to be in S?

(c) Since is transcendental, is it possible to have a non-zero polynomial p(x) with ?

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

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

Then you really just need to concentrate on 3 things:

(a) What does it

**mean**for to be an inverse of ?

(b) What does it mean for or to be in S?

(c) Since is transcendental, is it possible to have a non-zero polynomial p(x) with ?

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

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#18

(Original post by

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?

**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?

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

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...

**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...

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#20

(Original post by

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?

**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?

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