30Yeah. I proved the initial part in such a lengthy way that now my solution stinks in front of yours.
f(x+y)=f(x)+f(y).......
then show
f (x)=xf(1)........for all x.....
i got it 4 integer and rational....but how to proove it 4 real??
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23 Answers
62it says that
f(xn) -> f(x)
Not that
f(Σxn) -> Σ f(xn)
I think that is what you have done!
30
21yaar this is just cauchy's equation.. i knew the proof..actually latex is too good ;)
21We have f(x+y) = f(x) + f(y).This impliesf(x+ x+x...nth) = nf(x). So,for positive integer n , f(nx) = nf(x) Let x= \frac{m}{n}y. m,n are positive integers.We have nx = my ,
So f(nx) = f(my)\Rightarrow nf(x)= mf(y)\Rightarrow nf(\frac{m}{n}y) = mf(y) ,Thusf(\frac{m}{n}y)= \frac{m}{n}f(y). We have proved for all reals p and positive rationals q , f(pq) = qf(p) (*) When q is zero (*) is true. So (*) is valid for non-negative rationals.Further,in the given equation setting y = - x, we get f(-x)= -f(x). So,(*) is valid for all rational q. To prove this for all reals we need to assume that f is continuous.
62{xi} denotes a sequence...
this is a basic result from higher mathematics....
30"For every {xi} which tends to x as i goes to infinity". How? [7]
30Sir what does convergence exactly mean? As in when you say, {f(xi)} converges to {f(x)}, what does it mean?
62For every {xi} which tends to x as i goes to infinity, continuity implies that {f(xi)} will converge to f(x)
Where {ai} stands for a1, a2, a3........
30I couldn't understand anything about the continuity part. [2]
62no no {.} is used to denote a sequence..
because it is a set of many numbers.. one for each natural number...
11cant v do it like this--
Clearly the fn. is f(x)=x
therefore f(1)=1
therefore f(x)=x.1=x (for all x)
62no karna.. you are taking the convergence of the sum of a sequence...
Here you have any sequence.. simple ai not sum of ai..
3well i want to know whether the post no 3 is rong in neway.I dunno
nething abt convergence and divergence,i just want to know dat whether post #3 explains it for real values.
1thanx ,,,sir ......
i got..
√2=\lim_{n->infinity}a_{1}+a_{2}+a_{3}+...........+a_{n}
f(\lim_{n->infinity}a_{1}+a_{2}+a_{3}+.........+a_{n})=a1f(1)+a2f(1)...........+anf(1)=f(1)(\lim_{n->infinity}a_{1}+a_{2}+a_{3}+...........+a_{n} )
=f(1)√2
62See there is a definition of continuity which i would say is slightly (only) difficult..
which says that
If a function is continuous at x, then if any sequence xn converges to x then it means that The sequence f(xn) also converges to f(x)
It is also a fact that for every irrational number, you can construct a sequence of rational numbers which converges to that irrational number (Can you try to construct the sequence?)
Now can you try and think of how this works?
1@nishant sir for integers...,rational numbers..i have proved...but for all real numbers i am unable to understand the proof....in M.L khanna also the same words wer used as used by you....
i am unabe to understand this
" let{xn}be a sequence of rational number which represent x"..[2]what is the meaning of this....then it says
"since f is continous at x,the sequence {f(xn)} converges to f(x)".[2] ..how????.
and then they hav given the proof.......
3@xyz u can see i have done it for arbitrary x,if x is irrationational then also we have f(x)=f(x-1)+f(1)
62it is something that you should understand nicely...
To help you through the proof, try and prove that
f(nx) = n f(x) for +ve integers n
f(x/n) = f(x) / n (for all +ve integers n)
THen try to prove that
f(p/q x) = p/q f(x) for any integers p, q
Then sequence definition of continuity will lead you to the final conclusion....
1oh yes ...i forgot to mention that condition......it is continous for all x
66to prove it for all reals, we require continuity of f at atleast one point.