Proving Discontinuity at Integers for f(x) = [x2]-[x]2

[vkash]

first of all see question
question:-> Prove that f(x) = [x²]-[x]² is discontinuous for all integrals values except 1.
[x] denotes greatest integer function OR [x] means greatest integer less than or equal to x.

I think this is discontinuous for +- sqrt(k) where k denotes an whole number(0,inf).
try values you will get the point.

 

[SammyS]

What is your question?

 

[vkash]

What is your question?

Is this question correct or not. (red part)
OR
either this question correct or incorrect.(red part)
question is marked as red and my explanation is in black words.

 

[SammyS]

first of all see question
question:-> Prove that f(x) = [x²]-[x]² is discontinuous for all integral values except 1.
[x] denotes greatest integer function OR [x] means greatest integer less than or equal to x.

I think this is discontinuous for +- sqrt(k) where k denotes an whole number(0,inf).
try values you will get the point.

Yes, the question appears to be correctly worded.

You are correct in observing that it is discontinuous for additional values of x, not mentioned in the question.

As to whether the function is continuous at x = 1, what are the following limits?

lim _x→1⁺ f(x)

lim _x→1^‒ f(x)
​

 

[vkash]

Yes, the question appears to be correctly worded.

You are correct in observing that it is discontinuous for additional values of x, not mentioned in the question.

As to whether the function is continuous at x = 1, what are the following limits?

lim _x→1⁺ f(x)

lim _x→1^‒ f(x)
​

thanks now i am confirmed
If any other person wants to answer then he is welcomed.

 

[SammyS]

Evaluate f(x) on the interval 0 ≤ x < 1 and then on the interval 1 ≤ x < 1.4 .

 

[vkash]

Evaluate f(x) on the interval 0 ≤ x < 1 and then on the interval 1 ≤ x < 1.4 .

I also so do that but in between sqrt(25) to sqrt(36).

 

[SammyS]

So you agree that f(x) is continuous at x = 1 ?

 

[vkash]

So you agree that f(x) is continuous at x = 1 ?

Oh man(sorry if you female) you are getting wrong meaning of my words. I mean that it has also point of dicontinouty other than '1'.
first when i do this as i see this question i feel that it will discontinous at all integers after confirming it i try to put sqrt(k) where k is whole number[0,infinity) it was also discontinous at square root of all whole numbers. then i post it here to confirm that is it wrong and now i am sure that question is wrong.
Do you know any graph drawing software that support greatest integer function and fractional part function(x-[x]). I have already one but that does not support these 2 function({} and[]).

If you want to do a beautiful question then see this. I am not requesting for help for this question. This issue is solved after 64 replies of some good persons(do you have already seen that).
Just test your root basic with this question.

 

[SammyS]

Yes, as I said in post #4 of this thread, f is discontinuous for many values which are not integers. However, the problem only says to prove discontinuity at integer values other than 1.

So, it is not a wrong question.