Dear Numbers&Polys Enthusiast-Scholars

Problem A1:

Sinamon-David Lemma.  If
1:	x = x+x
then x = 0.

Proof: By adding the additive inverse of x to each side of (1),
	x - x	= [x + x] - x
		= x + [x - x] , by associativity of addition,
		= x + 0
		= x.
Hence 0 = x, as desired.  QED

Proof of A1:  The idea is to use distributivity, as follows:
	c·0	= c·[0+0] = [c·0] + [c·0] ,
by DMA.  Hence by the SimDav-Lemma, c·0 equals 0.	QED

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Proof of A2:
By uniqueness of additive inverse, ISTShow (It Suffices To Show) that
2:	b + [[-1]·b]  =  0
But note that
	0	= 0·b ,		by (A1) and commutativity of multiplication,
		= [1 + -1]·b ,	by AID
		= [1·b] + [[-1]·b] , by DMA.
Courtesy of our multiplicative-identity axiom, this yields
	0 = LHS(2),
as desired.  QED

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What follows is a rewording of Claire's proof.

Proof of A3:  Consider this inequality:

3:   |x+y|²	=<   [|x| + |y|]² .

Note that each of |x+y| and [|x| + |y|] is non-negative, courtesy
(1.20a) and (1.15a).  Consequently we can apply (1.16k) and (1.17a),
by having let

	a := |x+y|	and	b :=  |x| + |y| ,

to conclude from (3) that

4:  |x+y|	=<   [|x| + |y|]

holds, which is the triangle inequality.

SECOND STEP: Courtesy of (1.20c), we can restate our goal, (3),
thusly:

3':   [x+y]²	=<   [|x| + |y|]² .

Now
  LHS(3')	=   x²  +  y²  +  2·x·y .
Observe that
  RHS(3')	=  |x|²  +  |y|²  +  2·|x|·|y|
		=  x²  +  y²  +  2·|xy|  , by (1.20c) and (1.20f).

Courtesy of OA, then, our goal (3') reduces to

        2·x·y	=<    2·|xy| .

and hence, thanks to OM, to

3'':    x·y	=<      |xy| .

But this follows from (1.20d) applied to  a := xy .    QED

================ Here is Claire's proof, lightly edited: ====
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A3: PROOF:

     Firstly, |ab| >= ab
     by Thm 1.20d.

     So,     2|ab| >= 2ab
     by Thm OM  and using that 2>0 by Thm 1.5f.

     Thus a^2+[[2|ab|+b^2] >= a^2+[[2[ab]]+b^2]
     by OA twice.

     Consequently |a|^2+[[2|ab|]+|b|^2 >= a^2+[[2[ab]]+b^2]
     by Thm 1.20c.

     Note that |a|^2+[[2|a||b|]+|b|^2 >= a^2+[[2[ab]]+b^2]
     by Thm 1.20f

     This yields [|a|+|b|]^2 >= [a+b]^2
     by Thm 1.10b used twice.

     Consequently Thm 1.20c yields that
     [|a|+|b|]^2 >= |a+b|^2  .

     Therefore, |a|+|b| >= |a+b|
     by Thm 1.15k, since |a|+|b|>= 0 by thm 1.15a
     and |a+b| >= 0 by Thm 1.20a.
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Problem A4: Use the QF to find the zeros of the polynomial x²+8x+15.
Graph the poly and list the intervals where it is positive.

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Problem A5: Addition does NOT distribute over multiplication.
The other three statements are true.

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