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#1 2011-09-08 22:35:04

fromwodehouse
Member
Registered: 2011-09-08
Posts: 4

Proof

we had a test and i got these three numbers wrong because i really dont understand a single thing about it and im still in grade 9. and since it bugging me i want to know how to do these things. Thanks ahead! smile

                                         m-n
1)   Prove: if n<0 and |m|= ------   , -1<m<0 or m>1
                                         m+n   
                                       

                         4-x              x+3               2-3x
2)   Given m = -------  ,  n = -------  ,   p =  ------   , and m>n>p
                          3                  4                   5

                                                      7
  Prove: the possible value of x is  -    <x  <1
                                                     17


3)   Positive numbers a,b, and c satisfy the inequality a+b+c> abc. Prove that a^2+b^2+c^2>abc

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#2 2011-09-08 22:59:58

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,376

Re: Proof

Hi;

1 and 2 are totally unreadable on my browser. I am looking at 3.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#3 2011-09-08 23:08:59

fromwodehouse
Member
Registered: 2011-09-08
Posts: 4

Re: Proof

okay thanks, 1 and 2 have fractions, i'll try to re write it again smile

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#4 2011-09-08 23:15:27

fromwodehouse
Member
Registered: 2011-09-08
Posts: 4

Re: Proof

1)   Prove: if n<0 and |m|= m-n/m+n  , -1<m<0 or m>1

                                   
2)   Given m = 4-x/3,  n = x+3/4,   p = 2-3x/5  , and m>n>p
      Prove that the possible value of x is {-7/17<x<1}

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#5 2011-09-09 00:28:59

bob bundy
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Registered: 2010-06-20
Posts: 6,380

Re: Proof

hi bobbym

I'm seeing

1)   Prove: if n<0 and

-1<m<0 or m>1
                                     
and    2)   Given                             

                       

and m>n>p

Prove: the possible value of x is 


                   
Bob

Last edited by bob bundy (2011-09-09 00:33:28)


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#6 2011-09-09 00:39:13

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,376

Re: Proof

Hi Bob;

That looks right!


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#7 2011-09-09 00:46:44

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,380

Re: Proof

So now we've done the hard bit we  just need to do them. smile

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#8 2011-09-09 04:22:03

reconsideryouranswer
Member
Registered: 2011-05-11
Posts: 172

Re: Proof

fromwodehouse wrote:

1)   Prove: if n<0 and |m|= m-n/m+n  , -1<m<0 or m>1

                                   
2)   Given m = 4-x/3,  n = x+3/4,   p = 2-3x/5  , and m>n>p
      Prove that the possible value of x is {-7/17<x<1}

fromwodehouse,

because of the Order of Operations,
you must use grouping symbols for these:



Signature line:

I wish a had a more interesting signature line.

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#9 2011-09-09 20:09:06

bob bundy
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Registered: 2010-06-20
Posts: 6,380

Re: Proof

hi fromwoodhouse,

Here's my proof for (1)

Consider the case m ≥ 0

In that case treat |m| as just m

but we know from the case that m and 1 + m > 0 so

Now consider the alternative case that m < 0

That means we can replace |m| with -m

this time we know 1 - m > 0 as m < 0 so

If a product is negative then one factor must be + and one must be - so

either m > 0 and (m + 1) < 0 but this contradicts the assumption that m < 0

or m < 0 and (m + 1) > 0 which leads to -1 < m < 0

Now to look at number (2).  See next post.

Hope that helps

Bob

Last edited by bob bundy (2011-09-09 20:17:33)


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#10 2011-09-09 20:22:38

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,380

Re: Proof

hi again,

(2) Like this:

and

So put these together and you get what you wanted.

smile

Now for (3).  edit: Many hours later.  I cannot do this one yet.  I've put out a general request for more brains.

Bob

Last edited by bob bundy (2011-09-10 03:38:52)


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#11 2011-09-11 00:13:11

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,376

Re: Proof

Hi;

A little help.

3)   Positive numbers a,b, and c satisfy the inequality a+b+c>= abc. Prove that a^2+b^2+c^2>=abc

By Muirheads inequality:

Therefore it is not mandatory that this can be done by the AMGM.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#12 2011-09-11 06:29:28

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,380

Re: Proof

This is for grade 9 and the other two were fairly straight forward.  Could the question be:

Positive numbers a,b, and c satisfy the inequality a+b+c>= abc. Prove that a^2+b^2+c^2>=(abc)^2

because that is do-able.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#13 2011-09-11 06:38:06

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,376

Re: Proof

Hi Bob;

Positive numbers a,b, and c satisfy the inequality a+b+c>= abc. Prove that a^2+b^2+c^2>=(abc)^2

because that is do-able.

Not if you have

a = 2, b = 5 / 4, c = 2
a = 3, b = 1, c = 2

I am fairly sure the original problem #3 is true.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#14 2011-09-11 07:15:52

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,380

Re: Proof

Ok, my mistake.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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#15 2011-09-11 08:06:04

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,376

Re: Proof

It is okay. I am not sure about the problem either.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#16 2011-09-11 08:31:20

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Proof

hi guys

is this correct for (3):


“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

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#17 2011-09-11 08:42:03

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,376

Re: Proof

Hi;

This statement is not true under the conditions of the problem.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#18 2011-09-11 08:43:28

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Proof

why not?


“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

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#19 2011-09-11 08:46:21

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,376

Re: Proof

Hi;

why not?

Try a = 2, b = 2, c = 1.

I think this is where your error is.

You are okay after A)

Mistake made in B.

It was a pretty good idea though.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#20 2011-09-12 05:41:43

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,376

Re: Proof

Hi all;


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#21 2011-09-12 08:47:58

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,522

Re: Proof

what is muirheads' inequality?


“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

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#22 2011-09-12 09:22:34

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 86,376

Re: Proof

Hi;

I can use it a lot easier than explain it. Here is the site that taught me.

http://2000clicks.com/MathHelp/IneqMuir … ality.aspx


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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