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Topic:
Reason for operator precedence
Replies:
15
Last Post:
Mar 15, 2006 8:56 AM




Re: Reason for operator precedence
Posted:
Mar 14, 2006 1:13 PM


briggs@encompasserve.org wrote: > In article <1142344511.262841.322440@p10g2000cwp.googlegroups.com>, matt271829news@yahoo.co.uk writes: > > > > bri...@encompasserve.org wrote: > >> In article <1142342196.542632.294210@i39g2000cwa.googlegroups.com>, matt271829news@yahoo.co.uk writes: > >> > > >> > Tony wrote: > >> >> Hi all. > >> >> > >> >> Hope this isn't a silly question. > >> >> > >> >> I was wondering what the reason is for having multiple levels of operator > >> >> precedence? > >> >> > >> >> Phrased another way, why is it that we don't just evaluate everything from > >> >> left to right? > >> >> > >> >> Having multiple levels of precedence obviously adds complexity, so I assume > >> >> there must be some payback. However, I don't see what it is. > >> >> > >> > > >> > As far as addition/subtraction vs multiplication/division is concerned, > >> > one reason is to ensure that the distributive property of > >> > multiplication works sensibly. For example, we want 3*(4 + 6) = 3*4 + > >> > 3*6 = 3*(6 + 4) = 3*6 + 3*4. > >> > >> Remember that what we're talking about here is merely a notational > >> convention. It has nothing whatsoever to do with the distributive > >> property of multiplication over addition. > >> > >> You can express the distributive law for multiplication over division > >> using parentheses: > >> > >> a*(b+c) = (a*b) + (b*c) > > > > Obviously you can. I meant to make it work without needing parentheses, > > but it seems that wasn't clear. > > Ok. Try doing it using infix notation and the operator precedence > convention of your choice. Remember your rule: no parentheses > > Left to right doesn't work. > > b+c*a = a*b... and we're stuck > > Right to left doesn't work. > > b+c*a = ...b*c and we're stuck. > > Multiplication has precedence over addition doesn't work. > > a*... and we're stuck > > Addition has precedence over multiplication doesn't work. > > a*b+c = a*b+... and we're stuck > > Accordingly, trying to point to this case as a motivation for some > particular choice of operator precedence seems ill conceived. > > According to your argument, it follows that we are all using either > Polish (prefix) or Reverse Polish (postfix) notation.
Sorry, you've lost me. I was agreeing with you that even without any precedence convention we could still represent the distributive law as a*(b + c) = (a*b) + (a*c). However, the convention makes the parentheses redundant, because a*b + a*c is understood to mean (a*b) + (a*c). That's all I meant... possibly you are going into it at a deeper level than my simple observation warranted.



