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 Post subject: proof analysis
PostPosted: Sun, 30 Jan 2011 16:35:26 UTC 
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1) Prove that: if a>2 and b>1 ,then ab>2

2)State rhe axioms ,theorems or the definitions involved int he proof

3) State the laws of logic involved in the proof


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 Post subject: Re: proof analysis
PostPosted: Sun, 30 Jan 2011 16:36:53 UTC 
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idi wrote:
1) Prove that: if a>2 and b>1 ,then ab>2

2)State rhe axioms ,theorems or the definitions involved int he proof

3) State the laws of logic involved in the proof


What have you tried?

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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 02:10:35 UTC 
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Shadow wrote:
idi wrote:
1) Prove that: if a>2 and b>1 ,then ab>2

2)State rhe axioms ,theorems or the definitions involved int he proof

3) State the laws of logic involved in the proof


What have you tried?


Only the proof:

(a>2 and b>1) => (ab>2b and 2b>2) => ab>2


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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 02:48:36 UTC 
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idi wrote:
Shadow wrote:
idi wrote:
1) Prove that: if a>2 and b>1 ,then ab>2

2)State rhe axioms ,theorems or the definitions involved int he proof

3) State the laws of logic involved in the proof


What have you tried?


Only the proof:

(a>2 and b>1) => (ab>2b and 2b>2) => ab>2


Okay, how do you know it's legal to do that?

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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 03:00:42 UTC 
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Joined: Fri, 28 May 2010 02:57:26 UTC
Posts: 53
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
1) Prove that: if a>2 and b>1 ,then ab>2

2)State rhe axioms ,theorems or the definitions involved int he proof

3) State the laws of logic involved in the proof


What have you tried?


Only the proof:

(a>2 and b>1) => (ab>2b and 2b>2) => ab>2


Okay, how do you know it's legal to do that?


Because :

Since b>1 and 1>0 ,then b>0 and thus ab>2b

Also 2>0 hence 2b>2

So ab>2b and 2b>2,implying that ab>2


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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 03:03:23 UTC 
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idi wrote:
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
1) Prove that: if a>2 and b>1 ,then ab>2

2)State rhe axioms ,theorems or the definitions involved int he proof

3) State the laws of logic involved in the proof


What have you tried?


Only the proof:

(a>2 and b>1) => (ab>2b and 2b>2) => ab>2


Okay, how do you know it's legal to do that?


Because :

Since b>1 and 1>0 ,then b>0 and thus ab>2b

Also 2>0 hence 2b>2

So ab>2b and 2b>2,implying that ab>2


Right, now if you can just put names to those reasons--eg. the last one a>b and b>c implies a>c is called the transitive property--you will have the answer to the rest of the problem.

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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 03:15:10 UTC 
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Joined: Fri, 28 May 2010 02:57:26 UTC
Posts: 53
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
1) Prove that: if a>2 and b>1 ,then ab>2

2)State rhe axioms ,theorems or the definitions involved int he proof

3) State the laws of logic involved in the proof


What have you tried?


Only the proof:

(a>2 and b>1) => (ab>2b and 2b>2) => ab>2


Okay, how do you know it's legal to do that?


Because :

Since b>1 and 1>0 ,then b>0 and thus ab>2b

Also 2>0 hence 2b>2

So ab>2b and 2b>2,implying that ab>2


Right, now if you can just put names to those reasons--eg. the last one a>b and b>c implies a>c is called the transitive property--you will have the answer to the rest of the problem.


How about the laws of logic ??

Surely the transitive property is not a law of logic


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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 03:19:39 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
idi wrote:
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
1) Prove that: if a>2 and b>1 ,then ab>2

2)State rhe axioms ,theorems or the definitions involved int he proof

3) State the laws of logic involved in the proof


What have you tried?


Only the proof:

(a>2 and b>1) => (ab>2b and 2b>2) => ab>2


Okay, how do you know it's legal to do that?


Because :

Since b>1 and 1>0 ,then b>0 and thus ab>2b

Also 2>0 hence 2b>2

So ab>2b and 2b>2,implying that ab>2


Right, now if you can just put names to those reasons--eg. the last one a>b and b>c implies a>c is called the transitive property--you will have the answer to the rest of the problem.


How about the laws of logic ??

Surely the transitive property is not a law of logic


Most everything there is Modus Tollens or Modus Ponens.

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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 03:22:57 UTC 
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Senior Member

Joined: Fri, 28 May 2010 02:57:26 UTC
Posts: 53
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
1) Prove that: if a>2 and b>1 ,then ab>2

2)State rhe axioms ,theorems or the definitions involved int he proof

3) State the laws of logic involved in the proof


What have you tried?


Only the proof:

(a>2 and b>1) => (ab>2b and 2b>2) => ab>2


Okay, how do you know it's legal to do that?


Because :

Since b>1 and 1>0 ,then b>0 and thus ab>2b

Also 2>0 hence 2b>2

So ab>2b and 2b>2,implying that ab>2


Right, now if you can just put names to those reasons--eg. the last one a>b and b>c implies a>c is called the transitive property--you will have the answer to the rest of the problem.


How about the laws of logic ??

Surely the transitive property is not a law of logic


Most everything there is Modus Tollens or Modus Ponens.


How and where ??


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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 03:24:12 UTC 
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idi wrote:
How and where ??


If you cannot follow your own logic, then there's more of a problem than an analysis. Try and spot what you're doing, it's good to learn how you think.

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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 14:19:37 UTC 
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Joined: Fri, 28 May 2010 02:57:26 UTC
Posts: 53
Shadow wrote:
idi wrote:
How and where ??


If you cannot follow your own logic, then there's more of a problem than an analysis. Try and spot what you're doing, it's good to learn how you think.


How can one follow his own logic ?

Can you give an example??

How can somebody learn how to think??

Can you again give an example??


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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 14:54:48 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
idi wrote:
Shadow wrote:
idi wrote:
How and where ??


If you cannot follow your own logic, then there's more of a problem than an analysis. Try and spot what you're doing, it's good to learn how you think.


How can one follow his own logic ?

Can you give an example??

How can somebody learn how to think??

Can you again give an example??


I mean if I look at how I reason it's simple.

Code:
If A is a field, A is also a ring.

B is a field, hence B is a ring.


It's easy to see this is modus ponens.

So just look at each step, write out in words what your reasoning is. If you need to, you can also make a list of the logical inference rules then compare the forms of your reasoning sentences with the list and decide which rule you are using.

As for how someone can learn to think, that's a mystery even to the best psychologists. I recommend just reading a lot of proofs and getting used to how they read, and trying to write some yourself to mimic those. In this way you get practice learning what "good thinking" is like and attempting to duplicate it.

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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 21:27:05 UTC 
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Joined: Fri, 28 May 2010 02:57:26 UTC
Posts: 53
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
How and where ??


If you cannot follow your own logic, then there's more of a problem than an analysis. Try and spot what you're doing, it's good to learn how you think.


How can one follow his own logic ?

Can you give an example??

How can somebody learn how to think??

Can you again give an example??


I mean if I look at how I reason it's simple.

Code:
If A is a field, A is also a ring.

B is a field, hence B is a ring.


It's easy to see this is modus ponens.


This is not Modus Ponens

The general form of Modus Ponenes is:

[(p=>q) and p] => q

Is the above of that form??








Quote:
So just look at each step, write out in words what your reasoning is. If you need to, you can also make a list of the logical inference rules then compare the forms of your reasoning sentences with the list and decide which rule you are using.

As for how someone can learn to think, that's a mystery even to the best psychologists. I recommend just reading a lot of proofs and getting used to how they read, and trying to write some yourself to mimic those. In this way you get practice learning what "good thinking" is like and attempting to duplicate it.


honestly i make nothing out of the above


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 Post subject: Re: proof analysis
PostPosted: Mon, 31 Jan 2011 22:49:52 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
idi wrote:
Shadow wrote:
idi wrote:
Shadow wrote:
idi wrote:
How and where ??


If you cannot follow your own logic, then there's more of a problem than an analysis. Try and spot what you're doing, it's good to learn how you think.


How can one follow his own logic ?

Can you give an example??

How can somebody learn how to think??

Can you again give an example??


I mean if I look at how I reason it's simple.

Code:
If A is a field, A is also a ring.

B is a field, hence B is a ring.


It's easy to see this is modus ponens.


This is not Modus Ponens

The general form of Modus Ponenes is:

[(p=>q) and p] => q

Is the above of that form??

honestly i make nothing out of the above


This IS Modus Ponens. If you cannot recognize it, then you need some more practice with that. P is the statement (A is a field) Q is the statement (A is a ring).

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 Post subject: Re: proof analysis
PostPosted: Tue, 1 Feb 2011 01:34:54 UTC 
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Joined: Fri, 28 May 2010 02:57:26 UTC
Posts: 53
Shadow wrote:


Code:
If A is a field, A is also a ring.

B is a field, hence B is a ring.


It's easy to see this is modus ponens.

.


idi wrote:

This is not Modus Ponens

The general form of Modus Ponenes is:

[(p=>q) and p] => q

Is the above of that form
??


Shadow wrote:

This IS Modus Ponens. If you cannot recognize it, then you need some more practice with that. P is the statement (A is a field) Q is the statement (A is a ring).


If you put :

P = (A is a field) ...........Q = ( A is a ring).

Can we put then : P=(B is a field) and CONCLUDE ,(B is a ring) is a Q,by using M.Ponenes ??


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