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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 8:46 AM


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)
You can express the distributive law for multiplication over division using a convention in which * binds tighter than +:
a*(b+c) = a*b + b*c
You can express the distributive law for multiplication over division using a convention in which + binds tighter than *:
a*b+c = (a*b) + (b*c)
You can express the distributive law for multiplication over division using Polish notation:
*a+bc = +*ab*bc
You can express the distributive law for multiplication over division using Reverse Polish notation:
bc+a* = ab*bc*+



