]]>

I think you require to be positive as well.

Let

and . Then is convex for . Applying Jensen's inequalitygives

which is what's required.

]]>For real numbers a, b and c:

Prove that ]]>

Superlike!]]>

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 ifcondition)

Then you should know that for two real numbers it states that . Now set x=n and y=1/n.]]>

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)

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.

Do you know what the AM-GM inequality is?

]]>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.

]]>I will.

Hi anonynistefy;

Would you please explain the AM-GM ethod?

I rewrote the proof from post #2 in post #9 to make it more rigorous. Use that one.

]]>As for me post #2 was a hazy one, and I had doubts about it.

But from #9 I am now sure about the proof

Thanks all

]]>If a question contains elements of A,B and C and a strong candidate knows nine out of the ten topics really well you'd expect them to get high marks. After all, they know 90% of the work. But if, say C, is their weak point then they cannot do a whole question, even though they know A and B. So luck plays too big a part in the results.

I'll give you a real example that come up on a paper about 40 years ago. The candidates were given a diagram of a rhombus with the lengths of both diagonals. They had to calculate the length of a side, the perimeter, the area and the angles between the sides. That tests pythag, area of triangles, and trig but you have to know one crucial fact before you can do any of it. If you know that the diagonals bisect each other at right angles then you're away but without that fact you cannot do any of it!

Now certainly a candidate who doesn't know the properties of a rhombus deserves to be penalised, but maybe 1 mark, not 4. It puts too much weight on a single part of the syllabus which distorts the correlation between how much the students know and the mark they come out with.

Nowadays they try to set question parts which are independant of each other so that you are only penalised once for any lack of knowledge/skill. That's why they say 'show that'. It means you can still tackle the rest of the question and should do so! (using the thing you had to show)

I haven't heard that other subjects were affected. The media has been red hot on this so I think we would have heard if there was the slightest thing wrong elsewhere. You can ask for a photocopy of your paper but it costs money. And I think the deadline for this may have gone.

Your teacher will have the <real mark> to <UMS> conversion so you can see exactly how many marks you lost. Then you might be able to work out where you lost them.

Bob

]]>