MHB What Axioms Justify the Simplification of Polynomial Expressions?

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The discussion centers on the axioms that justify the simplification of polynomial expressions, particularly focusing on the removal of parentheses in the expression (x^2 + 2x + 5) + (x^2 + 3x + 1). Participants emphasize the importance of the associative and commutative properties of addition, which allow for rearranging and regrouping terms without changing the result. The distributive property is also highlighted as essential for factoring the final expression. The conversation reveals some confusion regarding the question's requirements, with a call for a clear list of axioms used in the proof chain. Ultimately, the axioms of real numbers, including the identity property, are crucial for justifying the simplification process.
paulmdrdo1
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in this problem we drop the use of parentheses when this step is justified by associative axioms. thus we write $\displaystyle x^2+2x+3\,\,instead\,\,of\,\,\left(x^2+2x\right)+3\,or\,x^2+\left(2x+3\right)$. tell what axioms justify the statement:

1. $\displaystyle \left(x^2+2x+5\right)+\left(x^2+3x+1\right)\,=\, \left(1+1\right)x^2+\left(2+3\right)x+ \left(5+1\right)$

i don't understand the question.
 
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paulmdrdo said:
in this problem we drop the use of parentheses when this step is justified by associative axioms. thus we write $\displaystyle x^2+2x+3\,\,instead\,\,of\,\,\left(x^2+2x\right)+3\,or\,x^2+\left(2x+3\right)$. tell what axioms justify the statement:

1. $\displaystyle \left(x^2+2x+5\right)+\left(x^2+3x+1\right)\,=\, \left(1+1\right)x^2+\left(2+3\right)x+ \left(5+1\right)$

i don't understand the question.
You first have to prove: (a + b) + c = a + (b + c) = a + b + c. (I'm assuming the final form is meant to suggest addition of the terms in any order.)

Then for problem 1 use the above result to remove the parenthesis, use commutivity of addition to rearrange the terms, then use the distributive property to factor.

-Dan
 
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why did you use associativity of addition?
 
i still don't understand what the question means.
 
paulmdrdo said:
i still don't understand what the question means.
I'm assuming that if the addition is associative and commutative then we can show
(a + b) + c = (a + c) + b = (b + c) + a ... = a + b + c because we can show that order doesn't matter. So we simply call it a + b + c.

The problem is asking you to use this to remove the parenthesis in the following:
(x^2 + 2x + 5) + (x^2 + 3x + 1) = x^2 + 2x + 5 + x^2 + 3x + 1

To get to the final form you can use commutivity to rearrange the terms, then use the distributive property to factor them to the final form.

-Dan
 
paulmdrdo said:
tell what axioms justify the statement:

1. $\displaystyle \left(x^2+2x+5\right)+\left(x^2+3x+1\right)\,=\, \left(1+1\right)x^2+\left(2+3\right)x+ \left(5+1\right)$

i don't understand the question.
The answer to this question should be a list of axioms. The axioms in question are used in a proof of the equality above. Roughly speaking, a proof in this case is a chain of expressions $E_1=E_2=\dots=E_n$ where each $E_i$ has some subexpression $e$, $E_{i+1}$ is obtained from $E_i$ by replacing $e$ with $e'$ and $e=e'$ or $e'=e$ is an instance of an axiom of real numbers. For example, a proof may start with \[(x^2 + 2x + 5) + (x^2 + 3x + 1)=(1\cdot x^2 + 2x + 5) + (x^2 + 3x + 1)\]Here $E_1$ is $(x^2 + 2x + 5) + (x^2 + 3x + 1)$, $e$ is $x^2$ and $e'$ is $1\cdot x^2$. The axiom used here is $1\cdot x=x$ for all $x$, and $1\cdot x^2=x^2$ is its instance.

So you need to list all axioms that are used in the chain of equalities \[(x^2+2x+5)+(x^2+3x+1)=\dots=(1+1)x^2+(2+3)x+ (5+1)\]
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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