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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: 90,580

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.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

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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
Moderator
Registered: 2010-06-20
Posts: 6,529

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: 90,580

Re: Proof

Hi Bob;

That looks right!


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Online

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

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,529

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
Moderator
Registered: 2010-06-20
Posts: 6,529

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: 6,529

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: 90,580

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.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Online

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

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,529

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: 90,580

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.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Online

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

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,529

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: 90,580

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.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Online

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

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,657

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: 90,580

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.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Online

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

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,657

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: 90,580

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.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Online

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

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 90,580

Re: Proof

Hi all;


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Online

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

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,657

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: 90,580

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.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

Online

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