Illidan2
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#1
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Prove that  n^2-n is an even number.

How is this deduced? What should I be thinking about in order to derive a sequence of logical steps leading to the solution?
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the bear
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#2
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(Original post by Illidan2)
Prove that  n^2-n is an even number.

How is this deduced? What should I be thinking about in order to derive a sequence of logical steps leading to the solution?
you can start by just putting in a few numbers to see how it works.

then think what you could do with the expression...
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RDKGames
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(Original post by Illidan2)
Prove that  n^2-n is an even number.

How is this deduced? What should I be thinking about in order to derive a sequence of logical steps leading to the solution?
You should factorise then inspect the parity of both numbers, thus logically deduce that the product MUST be even.
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Illidan2
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When I factorise  n^2-n I get  n(n-1) , do I not? So the parity of -1 is odd, but n can in itself be even or odd as n can be any number, right? I'm not sure where I should go from there, or if what I have done is even the logical form of progression one would make when trying to prove that  n^2-n .
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ArcaneMists
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Think about the relationship between n and n-1. When n is odd, what is n-1? When n is even, what is n-1?
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RDKGames
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(Original post by Illidan2)
When I factorise  n^2-n I get  n(n-1) , do I not? So the parity of -1 is odd, but n can in itself be even or odd as n can be any number, right? I'm not sure where I should go from there, or if what I have done is even the logical form of progression one would make when trying to prove that  n^2-n .
So you know that n can be either even or odd. You have 2 cases to consider. First, go with 'if n is even, then...' and arrive at a conclusion that the product must be even. Repeat similarly for the second case.
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ImprobableCacti
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(Original post by Illidan2)
When I factorise  n^2-n I get  n(n-1) , do I not? So the parity of -1 is odd, but n can in itself be even or odd as n can be any number, right? I'm not sure where I should go from there, or if what I have done is even the logical form of progression one would make when trying to prove that  n^2-n .
What is true about the parity of two neighbouring numbers?
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Illidan2
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When n is odd, n-1 is even.

When n is odd, n(n-1) is even.

When n is even, n-1 is odd.

When n is even, , n(n-1) is even.

I notice that in both cases of n(n-1), or  n^2-n , the result is even. I think I know how to prove this now. Give me a few minutes...
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RDKGames
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(Original post by Illidan2)
When n is odd, n-1 is even.

When n is odd, n(n-1) is even.

When n is even, n-1 is odd.

When n is even, , n(n-1) is even.

I notice that in both cases of n(n-1), or  n^2-n , the result is even. I think I know how to prove this now. Give me a few minutes...
You already proved it in this post though lol
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Illidan2
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Oh. I'm allowed to just write that as my proof? If I were to write something to that effect in an A-Level exam(naturally, I assume i'll be proving something more complex), would I receive the marks? Can I prove it in sentences? I thought I had to figure out some method of conversion into algebra that expresses my thoughts in a logical way. Am I overthinking it?
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RDKGames
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(Original post by Illidan2)
Oh. I'm allowed to just write that as my proof? If I were to write something to that effect in an A-Level exam(naturally, I assume i'll be proving something more complex), would I receive the marks? Can I prove it in sentences? I thought I had to figure out some method of conversion into algebra that expresses my thoughts in a logical way. Am I overthinking it?
Yes you can just write that down. There are different methods of proof to use and it often depends on the statement you're trying to prove. This one requires proof by cases. I don't think you'd encounter a proof by cases in an A-Level exam.

If you REALLY want to be more rigorous about it, you can say that in each case you have n=2m or n=2m+1 then in each case you can proceed to work with algebra and factor out a 2 to show that the product is even.
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Illidan2
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Oh, I see. Thank you so much! :3

Proof by cases(or exhaustion), is part of the specification for the A-Level Maths course I am working my way through, as well as deduction, counter-example and contradiction. Is proof by cases a derivative of proof by deduction? My textbook had that question listed as a "proof by deduction". I assume that because proof by cases is listed as part of the Edexcel A-Level Maths 2017 linear specification, it may be possible that I may encounter a question requiring me to use proof by cases?
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thekidwhogames
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Well an easy way to do it is notice these are consecutive numbers and therefore the product is even.
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ImprobableCacti
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(Original post by Illidan2)
Oh. I'm allowed to just write that as my proof? If I were to write something to that effect in an A-Level exam(naturally, I assume i'll be proving something more complex), would I receive the marks? Can I prove it in sentences? I thought I had to figure out some method of conversion into algebra that expresses my thoughts in a logical way. Am I overthinking it?
You could say that n(n-1) is the product of two consecutive numbers. Any two consecutive numbers will be a pair of odd and even numbers. The product of an odd and even number is always even (it will always have a factor of 2).
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RDKGames
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(Original post by Illidan2)
Oh, I see. Thank you so much! :3

Proof by cases(or exhaustion), is part of the specification for the A-Level Maths course I am working my way through, as well as deduction, counter-example and contradiction. Is proof by cases a derivative of proof by deduction? My textbook had that question listed as a "proof by deduction". I assume that because proof by cases is listed as part of the Edexcel A-Level Maths 2017 linear specification, it may be possible that I may encounter a question requiring me to use proof by cases?
If it was under 'Proof by deduction' then they'd probably want you to just immediately think "Oh it's the product of two alternative parities, so it must be even" without much maths going on like it would if it was under cases. If you can't deduce it, you can just use cases like you did here. (given that the proposition can be done by cases, of course)
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Illidan2
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#16
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I just checked the solution online. They deduced it by saying "If n is even, n-1 is odd and even*odd=even", and then repeated that for n being odd. Is that proof by deduction or cases?
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RDKGames
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(Original post by Illidan2)
I just checked the solution online. They deduced it by saying "If n is even, n-1 is odd and even*odd=even", and then repeated that for n being odd. Is that proof by deduction or cases?
Strictly speaking, it's cases. But deduction isn't its own way of proving a statement.
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Illidan2
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#18
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I see. Thank you!
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