Sergey Dukman

general recent conversations

$Mathematics$ Discussion between Thomas Prévost and…

Imported from a comment discussion on math.stackexchange.com/q...

Sergey Dukman

Jan 19, 2021 20:10

I just have another one question: How to show that the function is isometric?

Sergey Dukman

Jan 19, 2021 20:00

I understood. Thank you.

Sergey Dukman

Jan 15, 2021 08:10

Are you here ? :)

Sergey Dukman

Jan 14, 2021 08:33

Let x,y in R. f(x)=f(y) if and only if x=y. Try f(x) with x=2. f(2)=4. Thus to satisfy f(x)=f(y), f(y) must be equal to 4, f(y)=4. The only value that can satisfy this condition is y=2. Thus x=y=2 for all Natural numbers. I.e the function f(x)=x^2 is bijective for all Natural Numbers, and not bijective for the set of all Real numbers.
Is it correct?

Sergey Dukman

Jan 14, 2021 08:07

Oh, yes. By some reason I was thinking about positive numbers. Thus f(2)=4 and f(-2)=4. What does it says to us then?

Sergey Dukman

Jan 14, 2021 08:07

And in order to prove it we also need to take any positive numbers? Like f(1)=1 and f(4)=16?

Sergey Dukman

Jan 14, 2021 08:07

I guess that there is not such numbers that can satisfy f(a)=f(b)=4 for the given function.And since such numbers doesn't exist the condition of bijection can't be satisfied.

Sergey Dukman

Jan 14, 2021 08:07

Thank you for very scientific answer. Would you be so kind to show me on this example how it is done? Do I need to define the region R?

Sergey Dukman

Jan 14, 2021 08:07

Alright. Say we can have f(2)=4 and f(3)=9. Thus we have 2 different results 4 and 9 which are not equal and thus the function is not bijective. Is this correct? What is step 2? :)

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

Sergey Dukman

May 15, 2020 11:42

Can somebody take a look at my line integral problem posted on forum?