The Law of Ruling out Possibilities is emphasized in this chapter. It is used to eliminate each possibility one by one. The Who Owns the Zebra Puzzle is typical. This and other examples are available for extra credit. Formally it states: When p or q is true and q is not true, then p must be true.
The Law of Indirect Reasoning is perhaps better known when used in proof by contradiction. Here we note that something must be either true or false. We then assume the opposite of what we suspect to be the case. We then show that this leads to a contradiction. This shows that our assumption was not valid. These proofs often do not lend themselves to a two column format. However, it is important to make the structure and method of the proof clear. Thus these proofs should start with Suppose, Assume, or Either, continues until nonsense is obtain, which should be clearly labeled a contradiction, and close with a concluding remark regarding the validity of what we intended to prove.
We used this method to show that there are an infinite number of primes. The law formally states: If valid reasoning from p leads to a false conclusion, then p is false. We thus prove p to be true by assuming first it is false and showing this cannot be.
~(p v q) = ~p ^ ~q [The negation of (p or q) is a tautology of (not p and not q).]
~(p ^ q) = ~p v ~q [The negation of (p and q) is a tautology of (not p or not q).]
This is perhaps best demonstrated via a truth table.
When doing a coordinate proof, the figure is often drawn in a convenient location. The origin is often utilized for one vertex, with others located on the xaxis or yaxis. Symmetry is also exploited. Thus a rectangle might be located with vertices: (0,0); (a,0); (a,b); and (0,b), or (±a,±b). The last expression is considered ambiguous, but we mean all four possibilities here not just two. From the coordinates, one then determines lengths and angles.
Distance can be used to show that segments are congruent. Notice how the squaring makes it irrelevant which point is considered point 1 and which is considered point 2. Distance is also always considered to be a positive quantity. The distance formula also extends to three [or more] dimensions. However, the exponents remain 2 not 3 [or more].
D = sqrt[(x_{1}  x_{2})^{2} +
(y_{1}  y_{2})^{2}] D = sqrt[(x_{1}  x_{2})^{2} + (y_{1}  y_{2})^{2} + (z_{1}  z_{2})^{2}] 
The distance formula can be used to define circle and sphere algebraically.
circle: (x  h)^{2} + (y  k)^{2} = r^{2} 
sphere: (x  h)^{2} + (y  k)^{2} + (z  l)^{2} = r^{2} 
where (h,k) and (h,k,l) are the center and r is the radius. We will generalize the circle formula next year for other conic sections (quadratic relations) such as ellipses, parabolas, and hyperbolas. Remember also, that a lattice point is a point with integer coordinates. The twelve lattice points on the circle x^{2} + y^{2} = 25 are common on contests (and in problem 11.7#14).
We find the midpoint of a line segment with endpoints (x_{1}, y_{1}) and (x_{2}, y_{2}) similarly: ([x_{1} + x_{2}]/2, [y_{1} + y_{2}]/2). This also trivially extends into three (or more) dimensions: a line segment with endpoints (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) has midpoint: ([x_{1} + x_{2}]/2, [y_{1} + y_{2}]/2, [z_{1} + z_{2}]/2).
This, of course, is the arithmetic mean or average. We will deal with geometric mean later. Try proving this next theorem using a triangle located at (0,0); (a,0); and (b,c).
Midpoint Connector Theorem:
The segment that connects the midpoints of two sides of the triangle is parallel to the third side and half its length. 
Notice how the equation of a line in two dimensions is given by ax + by = c. Similarly, the equation of a plane in three dimensions is given by ax + by + cz = d. Can you predict the equation of a 3D space in four dimensions?
In two dimensions it is standard to graph the xaxis as increasing to the right and the yaxis as increasing upwards. Only occasionally will it differ with an axis reversed, the axes exchanged, or both. A rotation and/or reflection might be required to correct this.
The situation is far more complicated in three dimensions. Portraying three dimensions on paper is the first challenge. Many prefer to leave the x and yaxis as is and add the zaxis at a 45^{°} angle down through quadrant III signifying that it is coming out of the paper. Another common scheme puts the yaxis to the right, the zaxis up, and the xaxis coming out of the paper, again represented at an angle down through quadrant III. (I should note that it is common practice to shorten the scale on this axis by the square root of 2.) It is very important to label your axes, especially in three dimensions. Both of these representations obey a common convention called the righthand rule or righthand coordinate system. If you point your right fingers in the direction of +x, and curl them toward +y, your thumb points in the direction of +z. This important convention is embodied in the way many of our laws of physics are stated.
Additional vocabulary words: interpolation, extrapolation, incommeasurate, space (skew) quadrilateral
Example: Consider packaging constraints imposed by
some shippers, specifically, maximum length plus girth is
108", where girth is the perimeter of a cross section.
Assuming a square cross section, what is the maximum volume allowed.
Solution: Let 4x + y = 108 and V = x^{2}y.
Solve for y in terms of x, substitute into the volume equation,
take the derivative (V = 108x^{2}  4x^{3},
V' = 216x  12x^{2} = 12x(18  x) = 0),
set it equal to zero and solve for x (x=18; oops, that's calculus).
Instead, graph that volume equation V=x^{2}(1084x)
on the inverval 0 < x < 27 and 0 < y < 12000
and locate the maximum of 11664 in^{3} at 18" by 18" by 36"
using the 2nd calc feature of your graphing calculator.
Some useful mental math (and a tie in with algebra) comes up in this section, that is the rapid squaring of x+½, for example 8.5^{2}. Note how (x+½)^{2}=x^{2}+x+¼= x(x+1)+¼. Thus 8.5^{2}=8•9+0.25=72.25.
See also Verignon's Theorem: If the midpoints of consecutive sides of any quadrilateral are connected, then the resulting quadrilateral is a parallelogram.

The basic flow of an induction proof is as follows. First we demonstrate our statement to be true for n = 1. Next we demonstrate that if our statement is true for any n, then it is true for n + 1. Finally, we invoke Axiom 5 above and claim it is true for all n.
Example: Suppose we want a general formula to add up all the natural numbers
from 1 to n. We gave a formula back in Numbers Lesson 2,
but we didn't prove it, especially for odd n.
Solution: First we demonstrate our formula
T_{n}=n(n+1)/2
to be true for n=1.
T_{1}=1(1+1)/2=2/2=1 is the correct sum.
Now assuming T_{n}=n(n+1)/2
is true for n we see what happens when we add n+1
to it.
T_{n}+n+1=n(n+1)/2+n+1=
n(n+1)/2+2(n+1)/2=
(n(n+1)+2(n+1))/2=
(n+1)(n+2)/2=T_{n+1}
We see that when we add n+1 to T_{n} we
obtain the equivalent equation for T_{n+1}.
Since the statement is true for n=1,
and we have show that if it is true for T_{n}
it is true for T_{n+1}, by the induction axiom,
we have just proved it true for all n.
With this method it is fairly straightforward to derive formulae for the sum of squares and cubes as well. We will leave the associated algebra as homework. These will be useful in calculus when we take the limit of Riemann Sums to form integrals to find the area under curves.
The sum of i^{2} from i=1 to i=n is n(n+1)(2n+1)/6.
The sum of i^{3} from i=1 to i=n is (n(n+1)/2)^{2} = n^{2}(n+1)^{2}/4.
Thus BOB is wrong! Also, regarding that same problem, they seem to assume you fill the dome as well as the cylindrical part of the silo. You can go ahead and fill the dome portion today, but since silage settles so much, it won't be full tomorrow!
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