bob bundy wrote:

hi bobbym and all

Pedantic is what gets the money in math

Metaphorically ... maybe.  Literally ... not in my world.  I'd be a rich man. 

I've got another , even better example of why you shouldn't start with what you have to prove.

example.

Suppose 1 = 2

Then 2 = 1

adding

3 = 3

This is true.  So does that mean 1 = 2 ?

Bob

That would be just incarefullness. The steps are not equivalent.

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.

Last edited by noelevans (2012-10-07 13:33:47)

Agnishom wrote:

I know what AM-GM inequality is but have not made a lot of its use.

As for the logic I agree with bob bundy because If A implies B, then not necessarily B implies A. (As long as you dont prove the if and only if condition)

One more question:
For real numbers a, b and c:
Prove that

Last edited by Agnishom (2012-12-02 12:14:02)

Is there any other way than using derivatives?

Hi Agnishom;