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Algebraic Manipulation of Postfix Equations

Date: 05/14/2003 at 19:57:02
From: Aaron Stubbendieck
Subject: Algebraic Manipulation of Postfix Equations

Is it possible to solve a postfix equality algebraically, or does it 
need to be converted back to infix notation?

For example:
Infix: (5+4x)^5 = (2x-1)/3
Postfix: 4x(5+^5) = 2x(1-)3/

Because the order of operations is not directly associated with the 
affected terms, it is difficult to decide how to manipulate the 
components. If 2x is divided out, the result is 2(5+^5) = (1-)3/
Which cannot be converted to infix.

I've tried things like removing the operator that was required to 
make the change, in the example above, removing the / on the right 
side, but to no success. Is it simpler to just work with equations 
in infix form?


Date: 05/15/2003 at 17:05:26
From: Doctor Peterson
Subject: Re: Algebraic Manipulation of Postfix Equations

Hi, Aaron.

What a fascinating question! I don't think I've ever even considered 
doing algebra on postfix notation. In theory, surely it should be 
possible, since postfix is just a different way to say the same thing; 
but the rules will have to be stated in a different way, and may be 
more awkward. If nothing else, trying this should get us to think 
carefully about what we are really doing in algebra, rather than just 
following rules we've learned, which now seem "natural."

To begin with, we should use a different equation, because the one 
you chose can't be solved in infix form either. There is no general 
solution to fifth degree polynomials. I'd like to try first solving a 
slightly complicated linear equation, and then try a quadratic. If I 
forget, remind me I wanted to see what standard form for a polynomial 
looks like in postfix form, and how we can apply the quadratic 
formula as well as solving by factoring and by completing the square.

So let's solve this:

  4(x - 3) = 2x + 2

In infix form, we would first simplify (eliminating the parentheses 
and putting it into a standard polynomial form as a sum of terms), and 
then collect x's on one side and constants on the other:

  4x - 12 = 2x + 2

  4x - 12 + 12 = 2x + 2 + 12

  4x = 2x + 14

  4x - 2x = 2x + 14 - 2x

  2x = 14

  x = 7

Now let's put the equation into postfix form:

  x 3 - 4 * = x 2 * 2 +

Note that I didn't use a parenthesis as you did to indicate 
multiplication, but put in an explicit multiplication symbol. In 
postfix form you can't leave multiplication implied, and you can't use 
parentheses. Some other things about your postfix form suggest you 
don't have a solid grasp on how this form works, so let me know if you 
need more help with it.

There's one more thing we should consider: if we're to be consistent 
about postfix, shouldn't the equals sign be postfix, too? It can be 
considered as an operator that takes two expressions and returns as 
its value a true or false to indicate whether they are equal. Then 
our equation would look like this:

  x 3 - 4 * x 2 * 2 + =

This way, nothing is between the two things it refers to; so there are 
no left and right sides at all. That sure messes up the rules we're 
used to; and it makes it very hard to find where we want to work. 
There may be a way to do this, but at least for now I'd rather leave 
the equation itself as infix, so we can see two sides to it:

  x 3 - 4 * = x 2 * 2 +

Now we have to figure out how to simplify an expression. There are no 
parentheses to eliminate; but we'd like to have a standard form in 
which we multiply a variable by a coefficient, then add terms. So our 
standard form for a linear expression would be

  x a * b +

which corresponds to

  ax + b

How can we change

  x 3 - 4 *

into this form??

We'll need to learn the properties of addition and multiplication in 
postfix form:

                addition                   multiplication
                ---------------------      ---------------------
  commutative:  a b + = b a +              a b * = b a *
  associative:  a b c + + = a b + c +      a b c * * = a b * c *
  distributive:            a b + c * = a c * b c * +
  identity:     a 0 + = a                  a 1 * = a
  inverse:      a -a + = 0                 a 1/a * = 1
  inverse oper: a b - = a -b +             a b / = a 1/b *
                a a - = 0                  a a / = 1

Those give us some rules for manipulating an expression without 
changing its meaning. Just as we think of the distributive property as 
telling us we can move the multiplier inside the parentheses, now we 
will want to find ways to describe these manipulations so they are 
easy to remember. For now, I'll just have to refer to this list.

So to simplify our expression (which means, moving + to the right), 
we have to use the distributive property:

  x 3 - 4 * = x 4 * 3 4 * - = x 4 * 12 -

(Note that I had to work with a subtraction rather than an addition; 
that required me to justify what I did by writing a subtractive 
version of the property:

  (a - b) * c = a*c - b*c  ==>  a b - c * = a c * b c * -

Eventually, we will have to get used to postfix ways of thinking of 
subtraction as addition of a negative.)

Now I have this equation:

  x 4 * 12 - = x 2 * 2 +

We can next undo the last operation on each side (compare this with 
what I did before in infix):

  x 4 * 12 - = x 2 * 2 +

  x 4 * 12 - 12 + = x 2 * 2 + 12 +

To make the 12's cancel, I have to use the associative, commutative, 
and inverse properties together, which work this way:

  a b + b - = a b b + - = a b b - + = a 0 + = a

  x 4 * 12 - 12 + = x 4 *

The equation is now

  x 4 * = x 2 * 14 +

(Notice how I used the associative property to add the 2 and 12 on 
the right.)

Now we have to commute and subtract 2x from both sides:

  x 4 * = 14 x 2 * +

  x 4 * x 2 * - = 14 x 2 * + x 2 -

  x 4 * x 2 * - = 14

Then I combine like terms by factoring (distributive property in 
reverse):

  4 2 - x * = 14

  2 x * = 14

And at last I can commute and divide both sides by 2:

  x 2 * = 14

  x 2 * 2 / = 14 2 /

  x = 7

We did it! It looked pretty weird, didn't it? Yet I imagine if we were 
used to it, we could do it just as easily as we do it in infix now, 
because it would become familiar. And it's actually helpful that the 
last operation performed is always at the right, so it is natural to 
undo "from the outside in," which in infix form is a lot harder to 
recognize. That's the mistake you made in your attempt: you can only 
work on the right of an expression, not on the left.

Whether we could do this with the equality written in postfix is 
another question. On the whole, I think I'm happy to stick with infix 
notation.

If you have any further questions, feel free to write back. If you're 
interested in quadratics, I'll let you try it yourself first, and you 
can show me what you find.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra

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