9i dont know y u are repeatedly asking proof for such an evident thing byah
wen irrational
p/q <e +x ( x >p/q)
p/q > e+x (x <p/q)
have solutions
Let
x be a real number, and let e > 0. Then there
exist integers p and q > 0 such that
|x-p/q| < e
This might seem obvious! but try to prove this.. It is not very difficult either.
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31 Answers
62think of it like this
if x is a rational.. then the answer is obvious :)
x=p/q
now what can you say if x is irrational?
1one thing for x as negative just write x=-x and p = -p we need to prove the same almost now
1let us take first x as a positive real
now let us take p as n
then the real number x-n/q will be left of x ..
It has to cross the zero mark at some point :) so let the value of n such that the expression is nearest to zero and positive be N
then
x-N/q>0 but x-(N+1)/q <0
where as the differenvce between the two is just 1/q
so we have x-N/q <1/q
there are two cases
i) e>1/q
If this is the case then observe that x-N/q <1/q and positive too and hence we get p and q such that
|x-N/q| < e
ii) e<1/q
now as e is real e has to be greater than zero ! so e will be greater that 1/kq where k is an appropriate integer
then now we select again q as kq and do the same
now x-N/kq>0 and x-(N+1)/kq <0
so we have
x-N/kq<1/kq<e
thus as both are positive
|x-N/kq|<e
so in both cases we get p and q such that the following condition is satisfied
962well that is what we have to show
there exists a p and q such that
LHS is < RHS :P
1but ur q says |x-p/q| < e !!
how x=p/q ?
then e=0.
but u said e>0...
@!$#%^Q@##!!@$#!!!!!!!!!! [12]
21SINCERE REQUEST : as we approach the D'day of our lives, hame kyun zyada proofs 4 QOD can v pl. hav mor objectives, n kindof probables for these last few precious days. PL. sirJEE.............
NO DISRESPECT MEANT TO WONDERFUL PROOFS OF REAL MATHS!!
[1]
1|x- p/q| < e
1) [x-p/q] < e => p/q > (x-e)
p>0 , q>0 => p/q >0
if x<e, p/q ε (0,∞)
if x>e, p/q ε (x-e , ∞)
2) [p/q - x] < e => p/q < (e+x)
so, p/q ε (0 , e+x)
what more [12]
62celestine .. what do you mean by adjusted!!
That is what the question is..
you have to find the "adjustment" [3]
9if x ≥ e then p/q can be adjusted to make it anything less than e upto 0 also
if x<e then its obvious
62corrected the question .. that went wrong due to html formatting :P
1is it this or waht bhaiya ??? significance of e kya hai ?
\left|x- p/q \right|> e
24""for everyone who is dropping""...............
iska ishara meri taraf to nahin hai naa sir??????[9][9]
62now this should be tried..
for everyone who is dropping or in class xii...