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#1 2007-08-30 20:02:47

Daniel123
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Registered: 2007-05-23
Posts: 663

Simplifying

Quick question....

Just wondering, when a question says "simplify the following", which is more simple:

x (x + 1)

or

x² + x

and any reason(s) why?

Thankss big_smile

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#2 2007-08-30 22:49:16

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: Simplifying

I would say the first one because generally, factored expressions are simpler.


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#3 2007-08-30 22:51:40

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

Re: Simplifying

i think a better reason would be that each term in the expression is simpler.

x^2 is quadratic, whereas in x(x+1) both terms are linear


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#4 2007-08-31 00:04:08

Daniel123
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Registered: 2007-05-23
Posts: 663

Re: Simplifying

Rightt! Thanks big_smile

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#5 2007-09-01 02:04:55

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Simplifying

What counts as simple depends on the expression itself. For x[sup]2[/sup]+x and x(x+1), I’d say they are both equally simple.

Factorized expressions are not always the simplest. For example, if you factorize x[sup]2[/sup]+x−1, you get

Is this simpler than the original quadratic expression? I don’t think so. NoGood.gif

Last edited by JaneFairfax (2007-09-01 02:05:19)

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#6 2007-09-01 02:25:32

Identity
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Registered: 2007-04-18
Posts: 934

Re: Simplifying

But it's more useful big_smile

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#7 2007-09-02 01:10:04

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: Simplifying

Mmm, that's the main problem here, and why I was very careful to include a 'generally' in my first post.

I like to think of it as "how long it would take you to find the value when given an x", but as that's quite subjective it's still problematic.

There are exceptions, particularly when the expanded forms have cancelled terms.
For example, something that came up here before was (x+1)(x-1).

Normally, (x+a)(x+b) would definitely be simpler than x² + cx + d, but in this case (x²-1) there is no x term and so it gets harder to decide.


Why did the vector cross the road?
It wanted to be normal.

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#8 2007-09-03 03:42:22

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Simplifying

Simplicity can be measured in three different ways that I know of.  First, it can be measured in how long it takes you to write the thing out.  It can also be measured in by how long it would take a computer to compute it (counting the number of arithmetic operations).  Finally, it can be measured in by how well it makes certain things obvious.  For example, Jane's expansion of x2+x−1 would make it really easy to see that the root is plus or minus the golden ratio, which you can't get from x2+x−1.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#9 2008-07-25 08:22:28

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Simplifying

Yours truly wrote:

Factorized expressions are not always the simplest. For example, if you factorize x[sup]2[/sup]+x−1, you get

Is this simpler than the original quadratic expression? I don’t think so. NoGood.gif

Well, I’ve been studying some of the basics of Galois theory and I think I can offer a few remarks on this.

If we are considering the polynomial x[sup]2[/sup]+x−1 over

or
, that would be its “simplest” form. It cannot be factorized any further in
or
. But if it’s considered as a polynomial over
, then it can be factorized further in the manner shown above. Thus, whether x[sup]2[/sup]+x−1 is in its simplest form or whether it can be further simplified depends on which polynomial ring you are considering. smile

Technically, the polynomial x[sup]2[/sup]+x−1 is called an irreducible in

and
. The same polynomial is not an irreducible in
. The formal definition is this: If R is a commutative ring with multiplicative identity 1, an element u of R is an irreducible iff u ≠ 0, u has no multiplicative inverse in R, and whenever u = ab for any a, bR, it follows that either a or b has a multiplicative inverse in R.

Last edited by JaneFairfax (2008-07-25 12:03:58)

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#10 2008-07-25 10:16:54

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Simplifying

Galois theory is fun.  It really puts group theory in perspective.  If you've gotten to solvable extensions yet, trying proving that all p-groups are solvable.  That was probably my favorite proof in Galois theory because it uses a whole slew of things from group theory.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#11 2008-07-25 10:42:22

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Simplifying

Silly note on simplifying your example:




approx with numbers: 


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#12 2008-07-25 11:58:11

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Simplifying

Ricky wrote:

Galois theory is fun.  It really puts group theory in perspective.  If you've got to solvable extensions yet, trying proving that all p-groups are soluble.  That was probably my favourite proof in Galois theory because it uses a whole slew of things from group theory.

Well, I’m just starting on splitting fields now. smile I’ve just read a short chapter on constructible points – and it was most interesting to see how field theory could be used to prove that the angle 20° cannot be constructed using ruler & compass (or, at any rate, that the angle 60° cannot be trisected using only ruler & compass). big_smile

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#13 2019-11-11 18:11:52

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: Simplifying

This is how Phi is solved?


X'(y-Xβ)=0

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