2 mark proof question (1 Viewer)

oredbayz

New Member
Joined
Oct 9, 2021
Messages
29
Gender
Male
HSC
2021
RTP: 2^n + 3^n =/= 5^n

is it possible to do by induction? any help appreciated

thanks
 

notme123

Well-Known Member
Joined
Apr 14, 2020
Messages
1,002
Gender
Male
HSC
2021
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

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,113
Gender
Male
HSC
2006


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

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
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

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
Actually, proving the more general result that, for positive integers and , and for all integers


might have been easier.
 

5uckerberg

Well-Known Member
Joined
Oct 15, 2021
Messages
562
Gender
Male
HSC
2018
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

-insert title here-
Joined
Jun 19, 2014
Messages
2,556
Location
Outside reality
Gender
Male
HSC
2016
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

-insert title here-
Joined
Jun 19, 2014
Messages
2,556
Location
Outside reality
Gender
Male
HSC
2016
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....
 
Joined
Aug 26, 2020
Messages
5
Gender
Male
HSC
2021
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

Well-Known Member
Joined
Aug 5, 2021
Messages
783
Gender
Male
HSC
2020
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

New Member
Joined
May 11, 2020
Messages
3
Gender
Male
HSC
2021
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

New Member
Joined
Oct 25, 2019
Messages
3
Gender
Male
HSC
N/A
For n>1,
(2/5)^n + (3/5)^n < 2/5 + 3/5 =1.
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top