Prove that 2mn> mn as well as nm
Find the shortest proof.
HINT: this is in syllabus
oops sry, condition: n,m>0 and ε N
24yeah cele is rite...some cond is missing
WARNING:Long method.....(but did it just to check validity of question)
considering m,n>0
take log both sides
mn.log2>nlogm
=> n(mlog2-logm)>0
Since n>0 so (mlog2-logm) must be greater than 0
consider f(m)=p=mlog2-logm
p'=log2-1/m
=>p changes sign at m=1/log2
so the question should be true for m>1/log2 and n>0
11See Easy
2mn>mn
rite
Now you cancel n on both the sides
Therefore
2m>m
This is understandable therefore done
1wel...
to prove 2mn>mn
to prove 2m>m
to prove 2>m1/m
and isnt it obvious 2>m1/m????
in d same way....
to prove 2mn>nm
to prove 2n>n
to prove 2>n1/n........
12m>m
mlog2>logm
m>logm/log2
or by taking
2>m1/m
then take AM GM inequalities..
9it boils down to
2m>m
now m=1 is true
consider a function 2m -m f'(x) = ln2.2m - 1
this f'(x) > 0 for mε N
so its a strictly inc function
has m=1 is already true
its true for all mε N
1or.................
we cud do by PMI
2m>m
for f(1) 2>1 so f(k) is true
let f(k) be true so..
2k>k
now we need to prove f(k+1) is true
LHS..2k+1=2k.2
>k.2
>K+1
so f(k+1) is also true
so by induction...
2m>m
106easiest proof:
consider a set A having n distinct elements and set B having m distinct elements
No. of relations from A→B = 2mn
No. of functions from A→B = mn
As No. of relations < No. of functions so 2mn>mn
similarly taking A has m elements and B has n elements it can be proved that
2mn>nm
9but asish the human mind isnt framed to think in terms of #12
may be the Q setter worked backwards !!!!