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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;;;; End of file "a.soln.txt" ;;;;