Notes for AKT-140314/0:02:29 - Revision history

Drorbn at 18:51, 14 March 2014

2014-03-14T18:51:09Z

← Older revision		Revision as of 14:51, 14 March 2014
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	:In fact, I guess you can add to the list of allowed functions any continuous multivalued function with up to 4 values and it won't help (of course with 5 values you can solve it since the solution is such a function...). Am I right?		:In fact, I guess you can add to the list of allowed functions any continuous multivalued function with up to 4 values and it won't help (of course with 5 values you can solve it since the solution is such a function...). Am I right?

	I believe Boaz is right, and he is making a very strong point - not only there is no formula for the roots of a quadratic in terms of radicals - in fact this remains true even if we are allowed to use other ''univalent'' functions.		I believe Boaz is right, and he is making a very strong point - not only there is no formula for the roots of a quadratic in terms of radicals - in fact this remains true even if we are allowed to use other ''univalent'' functions (or even up to 4-valent).

Drorbn at 18:49, 14 March 2014

2014-03-14T18:49:12Z

← Older revision		Revision as of 14:49, 14 March 2014
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	Boaz wrote:		Boaz wrote:

	I have one comment- you claim it is weaker than Galois. At least in one		:I have one comment- you claim it is weaker than Galois. At least in one important aspect I think it is much stronger. I challenge you to show with Galois theory that if we add exp(z) to our list of allowed operations, a solution cannot be found...
			:In fact, I guess you can add to the list of allowed functions any continuous multivalued function with up to 4 values and it won't help (of course with 5 values you can solve it since the solution is such a function...). Am I right?
	important aspect I think it is much stronger. I challenge you to show with
	Galois theory that if we add exp(z) to our list of allowed operations, a
	solution cannot be found...
	In fact, I guess you can add to the list of allowed functions any continuous
	multivalued function with up to 4 values and it won't help (of course with 5
	values you can solve it since the solution is such a function...). Am I right?

	I believe Boaz is right, and he is making a very strong point - not only there is no formula for the roots of a quadratic in terms of radicals - in fact this remains true even if we are allowed to use other ''univalent'' functions.		I believe Boaz is right, and he is making a very strong point - not only there is no formula for the roots of a quadratic in terms of radicals - in fact this remains true even if we are allowed to use other ''univalent'' functions.

Drorbn: Created page with "Boaz wrote: I have one comment- you claim it is weaker than Galois. At least in one important aspect I think it is much stronger. I challenge you to show with Galois theor..."

2014-03-14T18:47:26Z

Created page with "Boaz wrote: I have one comment- you claim it is weaker than Galois. At least in one important aspect I think it is much stronger. I challenge you to show with Galois theor..."

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Boaz wrote:

I have one comment- you claim it is weaker than Galois. At least in one
important aspect I think it is much stronger. I challenge you to show with
Galois theory that if we add exp(z) to our list of allowed operations, a
solution cannot be found...
In fact, I guess you can add to the list of allowed functions any continuous
multivalued function with up to 4 values and it won't help (of course with 5
values you can solve it since the solution is such a function...). Am I right?

I believe Boaz is right, and he is making a very strong point - not only there is no formula for the roots of a quadratic in terms of radicals - in fact this remains true even if we are allowed to use other ''univalent'' functions.