Ok, thanks. So, within my answer, there is: if P then Q and if Q then P. so assuming P is true, since x<=1, this implies that Q is true. So to be more 'to the point' i should just write my proof in reverse with single headed implication arrows? this seems odd as joseph said, any ideas to clean it up?
My favorite way of doing these types of proof is by contradiction. So start with x+1−x>45 and assert that x≤1 and then work through until you reach something which can't possibly be true.
My favorite way of doing these types of proof is by contradiction. So start with x+1−x>45 and assert that x≤1 and then work through until you reach something which can't possibly be true.
thats a really good idea, thanks. It can be done using the same reasoning as well, so I think technically both proofs are correct, but this one is better
I just wanted to check my answers against yours. Don't do them if you don't want to.
Not sure if I went around it the right way. From the question, I was under the impression that the first word MUST be AB.
i) Start with AB. First rule (Ax -> Axx) gives ABB. Using the first rule again, ABB -> ABBBB. ii) I got AB^(2^n) where n belongs to the set [0,1,2..] iii) I worked backwards from the given to get a path of AB f-> ABB f-> ABBBB f-> ABBBBBBBB s-> CABBBBB s-> CCABB, where f-> means using the first rule and s-> means using the second rule. iv) Not ultimately confident that I'm expressing it properly, but I got all of C^(m)AB^((2^n)-3m), where n,m belong to the set [0,1,2..] and ((2^n) - 3m) >= 0. v) Caught me out I think. I couldn't understand why any combination of C^(n)AB^(m) couldn't occur, with n,m belonging to [0,1,2..], since the previous power of B that was 2n-3m was just to give the remaining number of Bs left at the end, but now they can be removed! So I leave my answer at C^(n)AB^(m), with n, m belonging to [0,1,2..]
I'd be interested to know what you got, or if you think I've gone about it the wrong way!
For the MAT, can we use methods that are NOT taught in the A Levels? Like Differentiation Under The Integral Sign? Wallis' Formula? Gamma functions and so on?
For the MAT, can we use methods that are NOT taught in the A Levels? Like Differentiation Under The Integral Sign? Wallis' Formula? Gamma functions and so on?
I don't see why not, but why would you want to/need to when the problems are simple enough not to require it?
For the MAT, can we use methods that are NOT taught in the A Levels? Like Differentiation Under The Integral Sign? Wallis' Formula? Gamma functions and so on?
I'm pretty sure you can answer the questions any way you wish to do so! Just yesterday, I read a post from the Oxford CompSci Dept. saying they would welcome an application from somebody who was part-way through a maths degree, but has realised they want to do CompSci - I'm sure he'd have a few ways of going about the MAT that most A-level students wouldn't!
For the MAT, can we use methods that are NOT taught in the A Levels? Like Differentiation Under The Integral Sign? Wallis' Formula? Gamma functions and so on?
I honestly don't recall having to use any of those things when doing MAT papers. Even in STEP, they're very rarely needed.
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Not sure if I went around it the right way. From the question, I was under the impression that the first word MUST be AB.
i) Start with AB. First rule (Ax -> Axx) gives ABB. Using the first rule again, ABB -> ABBBB. ii) I got AB^(2^n) where n belongs to the set [0,1,2..] iii) I worked backwards from the given to get a path of AB f-> ABB f-> ABBBB f-> ABBBBBBBB s-> CABBBBB s-> CCABB, where f-> means using the first rule and s-> means using the second rule. iv) Not ultimately confident that I'm expressing it properly, but I got all of C^(m)AB^((2^n)-3m), where n,m belong to the set [0,1,2..] and ((2^n) - 3m) >= 0. v) Caught me out I think. I couldn't understand why any combination of C^(n)AB^(m) couldn't occur, with n,m belonging to [0,1,2..], since the previous power of B that was 2n-3m was just to give the remaining number of Bs left at the end, but now they can be removed! So I leave my answer at C^(n)AB^(m), with n, m belonging to [0,1,2..]
I'd be interested to know what you got, or if you think I've gone about it the wrong way!
Thanks a lot. I stopped on ii), I'll do the rest tomorrow and see if I get something different.