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Tutors - All Subjects
AU:
tutorfinder.com.au
CDN :
findatutor.ca
CDN: .i-tutor.ca
CDN:
Montreal Tutors
NZ: findatutor.co.nz
UK:
tutorhunt.com
USA: wiziq.com
USA: ziizoo.com
tutor via them at your own risk. Good
luck.
YOU are better than YOU think. Show
yourself how:
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Read logic
chapters 1 to 5 in online volume Three
Skills for Algebra for greater skills & confidence
in work
and study.
Learn to read notes and textbooks like a lawyer, so that no nuance, no
subtlety and no clause escapes your attention |
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Logic
chapters 1 to 5 re- appear not in sequence, as is or longer,
in Volume 1A, Pattern Based
Reason, Bon Appetite.
Logic
Mastery
Amazing, Amusing, Amorous, Delicious, Delightful, Edifying,
Strengthening Elixir.
It eases work & learning difficulties Makes the hard easier. Opens eyes.
Leads to greater precision.
in reading and
writing
Logic
mastery makes the hard, easier. Logic
mastery leads to better, stronger and richer comprehension. Logic
mastery improves reading and writing. Logic
mastery ease learning difficulties. Logic
mastery gives a headstart. In sum, logic
mastery will develops critical thinking, improve reading and writing,
and give a firmer base for work and studies at many levels. Good luck.
After logic,
(a) continue reading Three
Skills for Algebra, chapters 8 to 14 and do so alongside site area on solving
liinear Equations ; or (b) see this calculus
starter lesson and Volume 3, Why
Slopes & More Math, chapters 2 to 6;
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Caution: Site advice is approximately
correct, for some circumstances, not all. That leaves room for thought |
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What may be learnt and when depends on how skills
and concepts are developed. Making the hard easier and clearer will allow
earlier & richer development of skills and concepts.
Try the Twiddla
Whiteboard. In principle, it allows
to people to draw and chat together online on a copy of this webpage or a clean
sheet. The chat may be via text or audio. Visit www.twiddla.com
to set up whiteboards to work with the webpage of your choice.
For online automated help in senior high school maths & calculus,
visit quickmath.com For Automatic
Calculus and Algebra Help with derivatives, integrals, graphs, linear equations,
matrix algebra, visit calc101.com
With overlap, each site quickmath
& calc101offers a different range of
services, some free, some not, all based on webmathematica. Good luck.
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Bolzano-Weierstrass Theorem
A set S is said to be infinite if and only if it contains an infinite
sequence of points qj (j=1,2,3, ...) with the property
that j ¹ k implies qj
¹ qk. A real number A
is said to be a limit point of a infinite set of numbers S if and only if
every interval I centered at A, no matter how small, must contain
infinitely many elements of the set S. Equivalently, a real number A
is said to be a limit point of a set S if and only if every interval I
centered at A contains at least one element s ¹
A of the set S.
Theorem B.1 [Bolzano-Weierstrass] If an infinite set S is
contained in a finite interval [a,b] then the set S has at
least one limit point A in the interval [a,b].
The main idea in the proof given next is as follows. For each whole
number k, in the finite interval [a,b] there must be a
leftmost subinterval Ik of length 10-k
which contains infinitely many points, which in turn must contain an leftmost
interval Ik+1 of length [1/2] 10-(k+1)
which contains infinitely many points, and so on. It follows that the left end
points of these nested (each inside its predecessor) intervals form a Cauchy
sequence with a limit L. This limit has the property that in every
interval of length [1/2]10-m about L,
there are infinitely many points of the infinite set S. Details follow.
Proof of the Bolzano-Weierstrass Theorem.
For each whole number k > 0, the interval [a,b] is
covered by subintervals, each of which has length 10-k.
We can choose these subintervals so that their end points have a decimal
representation that ends k places after the decimal point. If all the
subintervals contain only finitely many set members, the set S would be
finite. So at least one of the subintervals must have infinitely many elements
of S. While we don't have enough information to say which of the
finitely many subintervals must has them, there must be at least one
subinterval. Choose one, say the leftmost one, and call it Ik.
Now we have a sequence of intervals Ik. The above
choice implies that Ik+1 is contained within Ik.
The reason for this follows - a proof within a proof: The interval Ik+1
must be to the left, to the right, or inside of the interval Ik.
In the first case, Ik+1 would lie in an interval of
length 10-k which is to left of Ik
and has infinitely points, in particular, those in Ik+1.
But the latter is impossible because of the leftmost selection of Ik.
In the second case, the interval Ik contains 10
subintervals of length 10-(k+1).
But if Ik+1 is to the right of the interval Ik
then each of these ten subintervals contains only finitely many points of the
set. This contradicts the choice of Ik. The only
possibility that remains is that Ik+1 is contained
within Ik. That completes this proof within a proof.
The foregoing selection of intervals Ik yields, a
nested collection of subintervals, each of length 10-k
and each of which contains infinitely many elements of the original set.
End of proof. The above selection process implies that the left
end-point of the k-th subinterval has a finite decimal expansion 0.c1c2¼ck
which coincides with the first k places of the decimal expansion 0.c1c2¼ckck+1
of the left-end of the next subinterval. This process yields an infinite
decimal expansion 0.c1c2c3¼.
This defines a real number L = 0.c1c2c3¼
with the property that each neighborhood interval [L-10-k,L+10k]
contains infinitely many elements of the original set.
The conclusion (or assumption) that a bounded infinite set of real numbers
has a limit point has many consequences in arithmetic-based mathematics. The
above proof identifies the leftmost limit point of the set S.
| Here for the sake of contradiction, suppose B
< A is another limit point of the set S. Then A-B
> 10·10-k for some
whole number k. Further for such a whole number k, all
intervals of length 2·10-m
< 10-k centered at B
would also contain a point of S. And the latter would be imply
there was an interval Jk of length 10-k
containing B and to the left of A with infinitely many
points of S. But the decimal expansion of A to k-decimal
places, say Ak = c1c2¼cp.a1a2¼ak,
has the property that the interval Ik = [Ak,Ak+10-k]
is the leftmost interval with infinitely many points of S. Yet Jk
is to the left of Ik. This is a
contradiction. Thus the supposition that there exists a limit point B
of S with B < A must be false. |
Note the above arguments or reasoning depends on the assumption that an infinite
decimal expansions yields a real number. Base two or any other base m ³
2 could have been used instead. The selection of base ten in the above argument
is a historical and cultural preference.
FOOTNOTE: The set-theoretic formulation of modern math moves away from this
preference.
The above argument, that is proof, relies on the in principle ability
to choose subintervals, one inside another, repeatedly, one for each integer k
³ 1. The above demonstration (proof) is appealing,
but some could object to the use in-principle part of this argument. There is no
practical or constructive way to make the choice. That is, the choice is
possible in principle, but not in practice. For each whole number k, the
information that the set S is infinite is insufficient by itself to
identify in practice the leftmost interval of length 10-k
with an infinite number of points in S.
The above construction of a nested sequence of intervals is a plausible
argument for some, but only a figment of the imagination for others. There is a
division among mathematicians on whether or not thought-based but impractical
choice-based existence arguments (or constructions) are acceptable. The most
rigorous and also the most limited perspective is that the above kind of
argument is heuristic and somewhat plausible, but not reliable. Another
perspective is that the above argument is acceptable and reliable. Suffice it to
say that direct, not just in principle, existence proofs are more welcome and
more certain in the mathematical reasoning process than other kinds of proof.
However, some of the following theorems depend on the above theorem and hence
the above in-principle construction.
FOOTNOTE: More Food for thought: Compare and contrast the role of
choice in the above argument with the role of Maxwell's Demon in improbable
gas dynamics. There may be a limit to what is acceptable in principle as a
conclusion-reaching method.
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www.whyslopes.com
Real Analysis - Decimal View
Here are the Appendices from Volume 3, Why
Slopes and More Math, Chapters 14
to 19 in Vol 3 are related. Here is a reference for college or
university mathematics, electrical engineering and physics.
A. What's Next
B. Pigeon Hole Principle
B. Bolzano-Weierstrass
C1. Triangle Inequality
C2. Triangle Inequality
C. More T.Inequality
D. Sets & Sequences
D. Monotone Sequences
E. Limits, Properties
E Limits & Error Control
F. Continuous Functions
F. Closed Range Thm
F. Intermediate Val. Thm
F. Compactness Thm
F. Equicontinuity Thm
F Extreme Value Thm
G. Rolle's Theorem etc
G. Mean Val. Thm.
G. Constant Difference Thm
G. Lipschitz Continuity I
PS: One Sided Range Theorems
G. Velocity Revisited
G. Sufficient Conditions
H. Riemann Sums Conv
H. Lipschitz Continuity II
The site area More
Calculus contains a one-sided theorem with proof that should be of
interest too.
Vol 1A Logic Postscripts
online only:-
Proof
by Absurdity alias proof by contradiction
How
the demand for consistency supports the law of the excluded middle
Reality
versus or with the aid of Imagination
Science, Engineering & Math Students: Have you
seen a simpler geometric
introduction to complex numbers? ( java applet included) . Can you explain
what is a
variable without using a symbol? Can you derive trig
expression for dot & cross & cosine
law from complex number properties? For truth tables and indirect methods
of reason, see chapters
19-24 & postscripts in Pattern
Based Reason and visit Volume 1A, Pattern
Based Reason, striving for objectivity, the empirical challenge &
limits.
Vol 1A Postscripts
online only
Proof
by Absurdity alias proof by contradiction
How
the demand for consistency supports the law of the excluded middle
Help Me Learn/Teach;
- Algebra
words before symbols - direct
& indirect use of formula, numerical versus algebraic solutions - what
is a variable (more words)
- Arithmetic
- exercises
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videos on primes, lcm, gcm,lcd, square roots etc
- Calculus - geometric
preview, algebraic preview,
3 study guides,
much more
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-starter lesson with java applet - easy
consequences for trig & vectors in the plane
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Analytic - functions, polynomials, complex numbers, unit circle
trigonometry
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Direct & Indirect Use - Numerical versus Algebraic Solutions
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- Decimal View of concepts and of proofs
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