Unformatted text preview: ered a natural number. For our purposes it does not really matter.
All we need to know is whether we can count the natural numbers or not. One number more or less does not
make a difference. 8 p
/
Rationals: Q = { − p, q ∈ Z, q = 0 }
q
1. We know from class 4 Algebra between any two rational numbers there are
INFINITELY many rational numbers.
2. Even though there are infinitely many rationals between any two rationals
we can still COUNT or ENUMERATE the rationals in an orderly manner
without missing any. The rationals are COUNTABLE. This is how we may count
the rational numbers.
± 2/ → ± 3/
1
1 → → → → ± 2/
3 → ± 1/
4 → ± 1/
3 ± 4/
2 ± 5/ . . .
2 ± 3/
3 ± 4/
3 ± 5/ . . .
3 ± 4/
4 ± 5/ . . .
4 ± 2/ → ± 3/
2
2 ± 2/
4 → ± 1/
2 ± 4/ . . .
1 ± 3/
4 0/ → ± 1/
1
1 . . . and so on.
The set of rational numbers Q is more than DISCRETE.
We say that the set of rational numbers Q is DENSE.
Let us look again at our representation of numbers on an infinite line.
−3
−
2 −
3 −
2 −1
−
2 −
1 1
−
2 0 3
−
2 1 2 3 If we try to plot the rationals as colinear points in an orderly manner we would get
an “almost continuous”. The rational numbers by themselves do not exhaust,
continuous
COMPLETELY cover, the number line. 9 From the Set Theory point of view we see that N ⊂ Z ⊂ Q .
Can we say that all the POINTS on the line between the rational number
1 and the rational number 2 are rational numbers ? There are numbers that are not
rational numbers. Let us consider the square root of 2 denoted byS2 .
p
/
If S2 is a rational number then S2 = − p, q ∈ Z, q =0 .
q
Further, let p, q be such that they have no factor in common.
p2
2=−
q2
2q2 = p2
2q2 is even ⇒ p2 is also even ⇒ p is even.
Let p = 2n
Since 2q2 = p2 ⇒ 2q2 = (2n)2
⇒ q2 ⇒ 2n2 . 2n2 is even ⇒ q2 is even ⇒ q is even.
Thus p is even and q is even, implying they have as common factor 2. This contradicts
our condition that p and q have no common factor. Therefore...
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 Fall '09
 TAMERDOğAN
 Limit, Δx

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