Here are some basic lemmas that are used by the proof for Fermat's Last Theorem: n=5.
Lemma 1: (p+q)5 + (p-q)5 = 2p(p4 + 10p2q2 + 5q4)
(1) Using the Binomial Theorem:
(p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5
(p - q)5 = p5 - 5p4q + 10p3q2 - 10p2q3 + 5pq4 - q5
(2) Adding these two values together gives us:
2p5 + 20p3q2 + 10pq4 = 2p(p4 + 10p2q2 + 5q4)
QED
Lemma 2:
If:
t = q4 + 50q2r2 + 125r4
u = q2 + 25r2
v = 10r2
Then:
t = u2 - 5v2
(1) u2 = (q2 + 25r2)2 = q4 + 50q2 r2 + 625r4
(2) -5v2 = -5(10r2)2 = -500r4
(3) (q2 + 25r2)2 + -5(10r2)2 = q4 + 50q2 r2 + 625r4 + -500r4 = q4 + 50q2 r2 + 125r4
QED
Monday, October 10, 2005
Sunday, October 09, 2005
Some Simple Division Lemmas
In this blog, I want to outline some simple implications of the division which are very useful in basic mathematical reasoning:
Lemma 1:
If:
(a) c = a + b
(b) d divides a
(c) d divides b
Then:
d divides c
(1) By assumption (b) and (c), we know that there exists a',b' such that:
a = a'd
b = b'd
(2) So,
c = (a'd) + (b'd) = d(a' + b')
Lemma 2:
If
(a) c = a + b
(b) d divides c
(c) d divides a
Then
d divides b
(1) c = a + b → b = c -a = c + (-a)
(2) This results follows from Lemma 1 above.
Lemma 1:
If:
(a) c = a + b
(b) d divides a
(c) d divides b
Then:
d divides c
(1) By assumption (b) and (c), we know that there exists a',b' such that:
a = a'd
b = b'd
(2) So,
c = (a'd) + (b'd) = d(a' + b')
Lemma 2:
If
(a) c = a + b
(b) d divides c
(c) d divides a
Then
d divides b
(1) c = a + b → b = c -a = c + (-a)
(2) This results follows from Lemma 1 above.
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