Submitted by Anonymous (not verified) on
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For the second part of Q5, I did it slightly differently to the mark scheme. I want to ask if I'd still get credit.
In a similar manner to the integral maths solutions I said that 2^(1/3) and 2^(2/3) were either both rational or both irrational (from earlier), so set out only to prove the former was irrational.
So I said:
1)Assume 2^1/3 is rational.
2) Define S to be the set of integers for which n is a member IFF n*2^(1/3) is an integer.
3) Show S contains at least 1 member
4)Let k be the smallest element in S.
5) Show that (2^(1/3) - 1)k is in S.
6) 4&5 Yield a contradiction hence 1 is false.
7) Hence both 2^1/3 and 2^2/3 are irrational.
Then I showed 3,5,6 in exactly the same way as I did earlier.
Mark scheme not exhaustive
Submitted by cxm on
Hi, I'll have a proper look tomorrow, but the mark schemes are not supposed to be exhaustive (usually follow the method which is expected to be the most popular). Different correct approaches will also get credit and during marking the markers can consult the lead markers if they find anything unusual.
Following on
Submitted by cxm on
When you say you "said" that either both $2^{\frac 1 3}$ and $2^{\frac 2 3}$ are both rational or irrational, do you mean you proved that? you needed to show that if $2^{\frac 1 3}$ is rational (so can be written as $\frac p q$) then the other one is rational and if $2^{\frac 2 3}$ is rational then the first one is rational. However if you had done that then I can see nothing wrong with your argument (as long as you clearly explained hat you were using the previous result to shouw that $2^{\frac 2 3}$ is also irrational.