Math Is Fun Forum

Hi All!

Lugo should have put the steps in reverse order or at least mention that each step is reversible.
He has reversible steps in each of his examples.  This corresponds to each statement being
equivalent to the next.  This is logically the same as saying that if A,B,C,D are statements
then we have A <=> B <=> C <=> D.  This being the case, we can start with A and get D and
we can also start with D and get A.  But if we want A => D then to avoid being misleading to
others we should start with A and get D.  If one of these <=> breaks down; that is, only goes
one way, say B <= C, then we can only derive D => A but not A =>D.

Bob Bundy,  The example you gave of 1=2 thus 2=1 thus 3=3 is a great example.  The first
conclusion that 2=1 is  F  following from  F.   The second conclusion 3=3 is T following from F.  So it
is clear that from a False statement we can logically derive either True or False statements.  This
follows from the truth table for implicaton (=>).

  p   q   p=>q     
  T   T      T
  T   F      F       The implication being True corresponds to a VALID argument.
  F   T      T       The implication being False corresponds to an INVALID argument.
  F   F      T

Starting with a T statement the only thing that can be validly produced is other T statements, since
the second line indicates that starting with a T statement and arriving at a F statement means that
the implication (argument or step) is invalid.  Starting with a F statement we can validly arrive at
True or False statements as your example and the last two lines of the table show.

A variation of this problem is OFTEN seen in trigonometry.  Students trying to prove that A<=>F
come up with A =>B=>C and F=>E=>C and try to conclude that A<=>F is true because C<=>C
is true.  IF they had A=>B=>C=>E=>F they could conclude that A=>F,  BUT they DON'T have
C=>E and E=>F.  Instead they have E=>C and F=>E, the reverse implications (converses).
Hence they can't get from A to F validly. Again IF ALL the implications involved were DOUBLE
IMPLICATIONS (<=>) then all would be well.  They could go from A to F and also from F to A validly.

Pedantic?  Perhaps so, but then I am also a teacher and have seen this logical mistake over and
over when teaching trig.  A little bit of logic goes a long way in understanding mathematics.  Just
knowing the logic behind direct and indirect proofs and knowing how to negate statements (We
start ALL indirect proofs with the negation of the statement to be proven.) properly is critical to
being comfortable with doing proofs in mathematics (and elsewhere).

Why does the method of indirect proof work?   Because if we want to prove P to be true indirectly
we start with the  statement ~p (not P) and validly arrive at a false statement, then the statement
~P must be false (according to the second line of the truth table above) and hence P which is
equivalent to ~~P must be true.

Indirect proofs are quite nifty since they often allow us to start with more information than a direct
proof would and then what we must reach is "relaxed" in the sense that all we have to come up
with is ANY false statement (contradiction).  Indirect proofs are often much easier than direct
proofs and at times are the only known proofs for some theorems in mathematics.

Example:  To do a direct proof of  p => q we can assume p and then try to validly conclude q.
Or we can assume ~q and try to validly conclude ~P.  But for the indirect proof we assume
the negation of p =>q which is  p and ~ q, which gives us TWO bits of info to work with.  Then
all we have to do is come up with ANY false statement that we can.  When we validly arrive at
a false statement, the proof is immediately finished.

There are only seven simple rules for negations of statements and they are TOTALLY INDEPENDENT
of the CONTENT OR MEANING of the statements.  They are simply rules of symbolic logic.
1)  ~(p and q) is ~p or ~q    2) ~(p or q) is ~p and ~q     3) ~(p => q) is  p and ~q   
4)  ~( p <=> q) is either  p <=>~q  or  ~p <=> q  (your choice)
5)  ~~P is equivalent to p
6)  Change "for every"  to "there exists"  (When required) *
7)  Change "there exists" to "for every"   (When required) *

*  If a "there exists" or a "for every" statement is in the hypothesis of an implication then it is not
to be negated since the hypothesis p  of the implication  p => q is not negated in arriving at the
negation  p and ~q.   Likewise in negating a double implication you may or may not have to negate
depending on which of the two options in  4)  you choose.

As you might surmise, I am a great believer in teaching students a little bit of logic.  If they
continue in mathematics it could be a big boon to their understanding of proof and disproof.

Maybe my middle name should be "verbose."

Oops!  I got the T's and F's messed up in the  paragraph beginning with "Bob Bundy," so I hope
           this edit avoids any confusion.

Hi bobbym!

I've never seen the generalized "sum of roots" formula you gave in post#6.  That's really nifty!
Thanks for sharing it. 

Hi Agnishom! 

They have "programming languages" that allow you to write programs to explore math and
science using graphs, simulations, calculations (some quite massive) and whatever else you can
find to do with it.

Hi bobbym!  Top o' the mornin' to you!

Since I'm not very "functional" but more of a "procedural" type of individual I think that perhaps
I should  stick with Maple.  I'm having enough trouble trying to get a bit familiar with LaTex.

I've got too many "irons in the fire" to learn new languages unless I just can't get around it.  At my
age I probably don't have too many good years left so I'm wanting to make the most of them trying
to write articles about my passion (making math "user friendly") and putting them on my web site.

For many years I though that division was the only "unforgiving" operation out of +,-,*,/ in the
sense that any mistake in the calculation trashed everything following it.  But this spigot approach
doesn't have that problem so one can make a mistake in one place and not affect those following
just like for +, - and *.

Thanks for the input!  Have a super day today!

Hi bobbym!

Is use Maple to do most of my novice number theory calculations.  But it is limited to 268,435,448
digit numbers.  So it should be able to handle little problems like what is the 260 millionth digit after
the decimal in the division (10^900+3)/(2^3000+1), eh?  (I think it is a "5", but I wouldn't stake my
life on it!)

Searching on the internet, I couldn't find a spigot algorithm for these kinds of problems.  Have you
had any luck finding such? 

What mathematics package do you use for big number crunching?

Have a boukamendous day!

Hi zee-f!

In all these problems you are to find the supplement to the sum of the two given angles I assume.
That would be in all these cases 180 - (sum of the two angles).
Eg. 16)  180-(68+86) = 180-154 = 26.
      20)  180-(m+2m) = 180-3m.

Be blessed!

Hi tina123!

  Look at the number line:

   ... -3   -2    -1    0     1    2    3 ...
_____.___.___.___.___.___.___.___

To the left means smaller or less than.
Since -2 is to the left of -1, then -2<-1.

More generally

...-3<-2<-1<0<1<2<3...

You'll get it!

Hi SlowlyFading!

Here's another way of looking at the factoring of ax^2 + bx + c  ( a not zero) trinomials.  It is fully
equivalent to the FOIL method, but has a two dimensional set up rather than one dimensional.

The FOIL technique is equivalent to RCM (reversing cross multiplication for a 2x2 multiplication).

Here is the pattern:

.  .    .  .    .  .   where these three patterns from RIGHT TO LEFT produce
|        X        |   the units, 10', and 100's columns of a product in base 10.
.  .    .  .    .  .   In an unknown base x (polynomials) this is units, x's, x^2's.

Example:

        2  -3   Obtain the 3*(-4)=-12 from the RIGHT HAND pattern.
                  Obtain  2*4+3*(-3) = -1  from  the CROSS pattern.
        3   4   Obtain 2*3=6 from the LEFT HAND pattern.
        ____

   6  -1 -12   

As polynomials this is equivalent to (2x-3)(3x+4) = 6x^2 - x - 12.

Putting this "in reverse" we would be given the 6  -1  -12 and asked to find  the factors of this
which are  2  -3   and  3   4.

Ignoring the signs temporarily there are 6 possible ways to pair the factorizations ( 1*6 or 2*3 )
with the factorizations  ( 1*12 or 2*6 or 3*4 ).  Then we can switch the numbers in one of the
columns (but not in both at the same time)  which doubles the number of possibilities to 12. 

It's a matter of trial and error to find out which combination works (if any).  The more practice
one has doing this, the faster they can find the right one.

HELPFUL HINT for eliminating some possibilities:  If the product ( in this case the 6  -1  -12 ) has no
common factor > 1 (as is the case here) then NEITHER of the two factors we are looking for can
have a common factor > 1.  This eliminates lots of possibilities.  For example:  2   4  or  2   6
or  2   2  or  2  12  or  3   6  or  3   12  or   6   2   or   6    3   or   6   4   or   6   12  (ignoring
signs temporarily).   That's 10 of the 12 possibilities that can be eliminated without even multiplying
the two factors out.  All we have left is  (1   12  with  6   1) or (2   3  with   3   4).

Then since the rightmost term -12 in the product is negative we must make one of the numbers in
the rightmost column negative.  (If the leftmost column of the answer is positive (here 6) then it is
customary to use only positive numbers in the leftmost column although we could make them both
negative.)

The 1  12  with  6   1 would produce a middle term of  71 or -71 (considering the signs) and
the 2   3   with  3   4  would produce a middle term of 1 or -1.  The -1 is what we need.

So  (2  -3)*(3   4) produces the 6   -1   -12  as desired; that is, (2x-3)(3x+4)=6x^2 - x - 12.

My high school algebra book (from the 1940's) taught factoring trinomials of this type this way
instead of the FOIL setup.  I like it better and find that most of my students I show it to like it better also.

But it's a "take it or leave it" proposition.  Choose the one you like best. 

Hi Au101!

If one "counts" the points at 1 unit, 1/2 unit, 1/4 unit, 1/8 unit ... by simply passing across each of
these points on a number line (starting at 1 and proceeding towards zero) then it would take infinitely long to cross all these points IF there is a pause of 1 second at each point.  However if one walks a CONSTANT RATE of 1 unit per second going from 1 to 0, then it takes only one second to "cross" all these points and hence finish "counting" all the positive integers.  So it takes hardly any time at all (an infinitesimal amount of time) to "count" a point by this scheme.  Of course "counting" the points simply means to put them in a 1 to 1 correspondence with the positive integers, which is done theoretically without consideration of time.

So the reason most folks have a problem with this "counting" is that they don't take into consideration the time element perhaps intuitively thinking that there must be some pause when
counting each point.   Hence they think of it as a paradox. 

I guess mathematicians "plink away" or "plunk away" at Planck's constant to some extent.

Ah! isn't math wonderful?

Hi tina 123!

Start with 1 and add 1 to that at the end of the 1st second to get 2.  Then add 1 to 2 at the end of one half second to get 3.  Then add 1 to 3 at the end of 1/4 second to get 4.  Then add 1 to 4 at the end of 1/8 second to get 5.  Continue this for a total of two seconds.  Then you have counted all the natural numbers by the end of second number two.  To reach and count the number k it occurs at the end of the 1/(2^(k-2)) time period.  The "sum" of all of these time periods is two seconds, so by the time two seconds passes doing this process we have counted all the natural numbers.

If someone says, "But no, we didn't count the number, say m, then we counter that at the end of the 1/(2^(m-2)) time period we counted m.  So there were no natural numbers left out of the counting.  1 + 1/2 + 1/4 + 1/8 + 1/16 + ...  = 2  (That is, the limit is 2).  But 2 is a finite number so we can "continue this process for two seconds" to count all the natural numbers.

Infinity is a strange thing!  Who can understand it?  There are a number of interesting paradoxes dealing with infinity.

Hi SlowlyFading!

Clicking on your LaTex I think I can see what the problems look like:

1)  (21x^3+14x)/7  =  21x^3/7  +  14x/7  =  ?

2)  (3(2x-3)-x(2x-3))/(2x-3)  =  3(2x-3)/(2x-3)  -  x(2x-3)/(2x-3)   =  ?

3)  ((x^2(5x+6)-3(5x+6))/(5x+6)  =  x^2(5x+6)/(5x+6)  -  3(5x+6)/(5x+6)  =  ?

In each of these we are dividing the denominator into each of the two terms.  Then in
each of these cancel to obtain the final answer.

Or you can take the original numerators and factor them and then cancel:

1)  7x(3x^2+2)/7   =  ?

2)  (2x-3)(3-x)/(2x-3)  =  ?

3)  (5x+6)(x^2-3)/(5x+6)  =  ?

The first approach is probably easier.

Can you take them from here? 

Hi bobbym!

Finally!  Patience works.  I now have a signature.  I'll probably change it to something else before
long since I just put something in the "slot" to see when it would finally appear.

Have a lovely day!

Hi Grader!

Graphically this would be a point on the unit circle: one unit in the direction of positive 1rad or about 57.3 degrees.

Have a great day!