Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

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

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!

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

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

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

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

Offline

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

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}

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

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)*

Children are not defined by school ...........The Fonz

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

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Bob;

That looks right!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

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

Bob

Children are not defined by school ...........The Fonz

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

Offline

**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 171

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.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

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)*

Children are not defined by school ...........The Fonz

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

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

hi again,

(2) Like this:

and

So put these together and you get what you wanted.

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

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

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

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,340

Ok, my mistake.

Bob

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

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

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

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

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

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

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

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi all;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,037

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

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

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline