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2 mark proof question (1 Viewer)

oredbayz

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RTP: 2^n + 3^n =/= 5^n

is it possible to do by induction? any help appreciated

thanks
 

notme123

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it was 2 marks and it didnt ask induction. i think theres another intended method. would anyone happen to have a solution that is equaivalent to 2 marks
 

Trebla

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The middle term is non-zero/positive provided n>1 as an integer so hence they cannot be equal.
 

CM_Tutor

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I wonder what they would make of a proof that invoked Fermat's Last Theorem and noted that this was proven by Andrew Wiles a few years ago, pointing at that if there are no solutions in positive integers , , and for the statement for any integer , and then pointing out that with , , and , the case is also false as . Thus for all integers .
 

CM_Tutor

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Actually, proving the more general result that, for positive integers and , and for all integers


might have been easier.
 

5uckerberg

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Will there be anybody here showing Sydney Morning Herald how to do Q16 this year?

I am curious. There will always be those boomers who say so what, skills in the real world are more important than a little mathematical equation. Typical boomer talk from the Sydney Morning herald. They never learn.
 

Paradoxica

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For a non-binomial method, divide both sides by 3^n to obtain

(2/3)^n + 1 = (5/3)^n

The left hand side has a base smaller than 1 and is a strictly decreasing function of n, and the right hand side is likewise strictly increasing.

Thus, there can only be at most one solution, which is obviously n=1. Hence, the two expressions are unequal for n>1
 

Paradoxica

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I wonder what they would make of a proof that invoked Fermat's Last Theorem and noted that this was proven by Andrew Wiles a few years ago, pointing at that if there are no solutions in positive integers , , and for the statement for any integer , and then pointing out that with , , and , the case is also false as . Thus for all integers .
1995 is more than a few years....
 
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I let the LHS and RHS be different functions then I differeniated both functions and showed that 5^x increases faster than 2^x
+ 3^x after x =1. Dunno if its worth full marks but it worked out for me.
 

icycledough

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I let the LHS and RHS be different functions then I differeniated both functions and showed that 5^x increases faster than 2^x
+ 3^x after x =1. Dunno if its worth full marks but it worked out for me.
That does seem like a valid way of going about it (I think there must be various ways to go about it, so as long as you could set your answer out and it made sense in a chronological manner, then it should be fine). I guess with the topics of proofs, what makes it nice is the level of flexibility with regards to what method you can use for an individual question.
 

epicuber7

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I factorised 2^n + 3^n as (2+3)(2^(n-1) - 3*2^(n-2) + ... + 3^(n-1)) and assumed the right hand bracket could not be 5^(n-1) rip
 

dwhinc

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For n>1,
(2/5)^n + (3/5)^n < 2/5 + 3/5 =1.
 
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