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**Prove this**

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

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.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

Hi Agnishom

Do you know what the AM-GM inequality is?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

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 (2012-10-06 14:33:47)*

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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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)

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

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 ifcondition)

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

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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Oh! Great!

Superlike!

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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One more question:

For real numbers a, b and c:

Prove that

*Last edited by Agnishom (2012-12-01 13:14:02)*

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**scientia****Member**- Registered: 2009-11-13
- Posts: 224

I think you require to be positive as well.

Let

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

which is what's required.

*Last edited by scientia (2012-12-01 15:47:21)*

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Is there any other way than using derivatives?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**scientia****Member**- Registered: 2009-11-13
- Posts: 224

What derivatives?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,715

Hi Agnishom;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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