24kya galti???
its for postive integers..
1) The largest positve integer is 1
Surprised ????
well its true............
here is the proof....
Let n be the largest postive integer...
Also we know that n2≥n
Now since n is largest => n2=n
=> n=1
Hence proved...[3][3][9][9]
And plzz all those who know the source of these..dont reveal..[1][1][6]
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16 Answers
24SInce n is largest..so will we have any integer bigger than it ????SO its obvious that n will only be n2
23n2 ≥ n ..... (where n ≡ N)
i.e n2 > n ...(for n ≡ N, n>1)......(1)
and n2 = n for( n =1 ) ..(2)
now if we wanna talk about largest integer .....den we will
obviously consider only eqn (1) and not eqn (2) ......bcz .. eqn 2 is for those integers wich are gr8er dan 1 .....
hence the largest integer satisfies eqn 1 and not eqn 2
i think this is d trick ..isnt it ?
3lolz can i say onething ∞2=∞ which also satisfies infact this only satisfies if n is large lolz.Or we can say onething that the roots of n2-n=0 is n=0,1,∞
24btw here is new
2) For n>0,
(n - 2)2 = n 2
That means every second square is same...sounds strange....here is the proof
We do it by PMI
let us denote the statement by S
Step1)
Now S1 is obviously true
Step2)
Assume Sk is true
Step3)
Prove for n=k+1 =>Sk+1
Putting n=k+1 in S we get,
(k-1)2=(k+1)2.
Expand to obtain k 2 - 2k + 1 = k 2 + 2k +1
=>-4k = 0
multiply by (1 - 1/ k) (becoz k > 1)
=> -4k + 4 =0.
Add k 2 to both sides
=> k2 -4k +4 = k2
=>(k - 2)2 =k 2
Since Sk+1 reduces to the known statement Sk
=>Sk+1 is also true.
So Sn is true for all n > 1.
Come on ..can u spot the flaw ??
1i think you have proved -4k = 0 or k = 0 which is not possible as k>1
1well....if -4k = 0
=> k=0
but it is known that k>1
therefore, there is a contradiction
62Eureka.. you have proved that the largest integer is 1 if it exists...
You have not proved that the largest integer is 1
The subtle difference is this... You have assumed n as the largest integer.. which you cannot always :D
I mean your assumption that n is the largest integer means that you have assumed that such an integer exists which may not be true at the first place
62[1]
Where did you get this one from? I mean this was interesting since I had never seen such a reasoning !
49lekin bhaiya isme itna uchal ne ka kya hai.......sorry no offence....but....
he said n2≥n
but according to my logic the inequality has nothing to do with the question........since this is true for all integers and also the greatest one and the smallest one too........
62but since n is the largest integer and n^2 is greater than equal to n
there cannot be an integer (n^2) which is greater than n
so n^2 is equal to n