Re: How to factor a rational
- To: mathgroup at smc.vnet.net
- Subject: [mg78856] Re: How to factor a rational
- From: Steven Siew <siewsk at bp.com>
- Date: Wed, 11 Jul 2007 06:12:28 -0400 (EDT)
- References: <f6vntj$ql7$1@smc.vnet.net>
Thanks to everyone who reply to me via email.
What I was looking for is the way to do a rule replacement to extract
out the rational.
Thanks to your help, I'm able to understand how to do it. For the
record , this is what
I have learned.
$ cat file7.out
Mathematica 5.2 for Students: Microsoft Windows Version
Copyright 1988-2005 Wolfram Research, Inc.
In[1]:=
Out[1]= {stdout}
In[2]:= (* Write your mathematica code below *)
In[3]:= eq1 = (4/3)*L*A - (4/3)*L*B == a*k*t
4 A L 4 B L
Out[3]= ----- - ----- == a k t
3 3
In[4]:= FullForm[eq1]
Out[4]//FullForm=
> Equal[Plus[Times[Rational[4, 3], A, L], Times[Rational[-4, 3], B, L]],
> Times[a, k, t]]
In[5]:= eq2 = (4/3)*L*(A - B) == a*k*t
4 (A - B) L
Out[5]= ----------- == a k t
3
In[6]:= FullForm[eq2]
Out[6]//FullForm=
> Equal[Times[Rational[4, 3], Plus[A, Times[-1, B]], L], Times[a, k, t]]
In[7]:= eq3 = eq1 /. HoldPattern[Plus[Times[Rational[n_,m_] ,a_,b__] ,
Times[Rational[p_,m_] ,a_,z__]] /; n == -p ] :> Rational[n,m] * a *
(b - z)
4 (A - B) L
Out[7]= ----------- == a k t
3
In[8]:= FullForm[eq3]
Out[8]//FullForm=
> Equal[Times[Rational[4, 3], Plus[A, Times[-1, B]], L], Times[a, k, t]]
In[9]:= (* End of mathematica code *)
In[10]:= Quit[];