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Mathematical Proof

I assumed the commutative law of multiplication when I proved the Big
Distributive Law.
This allowed me to use the left distributive law alone.
(When you have commutativity, you only really need one distributive law.)
Don’t assume commutativity. Redo the proof so that it works with noncommutataive algebra, such as matrix algebra.
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Describe a strategy for computing the left side in a computer language that won’t allow you to throw nested loops on the fly.
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The proof the Big Distributive Law completes the proof of the
Fundamental
Theorem of Arithmetic.
The FTA gives us many irrationality proofs.
How about sqrt{3} – sqrt{2}.
T = sqrt{3} – sqrt{2}
T^2 = 3 – 2sqrt{6} + 2
(5 – T^2) / 2 = sqrt{6}
This if T were rational, sqrt{6} would be rational.
But the proof for the sqrt{2} works for sqrt{6}.
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(sqrt{3} – sqrt{2})(sqrt{3} + sqrt{2}) = 1
Thus, if either of the quantities sqrt{3} – sqrt{2} or sqrt{3} +
sqrt{2}
is rational, so is the other. Then, so is their sum and difference.
So that would make both sqrt{2} and sqrt{3} rational, contradiction
either way.
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Multiply out the polynomial
(x + sqrt{3} + sqrt{2})*(x + sqrt{3} – sqrt{2})
*(x – sqrt{3} + sqrt{2})*(x – sqrt{3} – sqrt{2})
Conclude that this is a polynomial with integer coefficients and
leading coefficient
equal to 1.
A polynomial with integer coefficients and leading coefficient
equal to 1 is called a **monic polynomial**.
Claim: The roots of a monic polynomial are either integers or
irrational.
This applies to sqrt{2} (root of x^2 – 2)
to cube root of 2 (root of x^3 – 2)
and by the exercise, to sqrt{3} – sqrt{2}
…and to much else besides.
Fix a monic polynomial p(x) = x^n + sum_{i=0}^{n-1} a_i x^i.
Without loss of generality, assume that a_0 not= 0.
(If not, divide out some x’s.)
Now assume for contradiction that s/t is a root of p: p(s/t)=0 and |
t|not=1.
(Assume s/t is in lowest terms.)
Consider the expression t^n * p(s/t). The value of this expression
equals 0,
by the assumption. Let Q be a prime that divides t.
t^n * p(s/t) = t^n * ((s/t)^n + sum_{i=0}^{n-1} a_i (s/t)^i)
= s^n + sum_{i=0}^{n-1} a_i s^i*t^{n-i}
Every one of these summands is divisible by Q except the first.
…flesh out the details and finish the proof.
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You learned in HS that rational numbers have repeating decimals and
vice versa.
1/7 = .
142857142857142857142857142857142857142857142857142857142857142857…
. 1 4
________________________________________________________
7 ) 1 . 0 0 0 0 0 0
7
—–
3 0
2 8

2

:
Repeating decimals for rational numbers is a special case of a fact about computations you can do with finite state machines
(with no input)
The denominator gives me an upper bound for when the repeating will start.
It also gives me a bound on the length of the cycle.
HOMEWORK: Divise a notation so you can turn this into a real proof.

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