At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

1^1 +1^0 = 1^1^1
2=1 which is FALSE

So, n=1 is not a solution

Should I post my solution?

Oh,,Sorry about that.

As u say

2 minutes, I'm having a crack.

U can take your time

Apologise if it is obvious to have solutions, its been a long day of uni

Actually, NO SOLUTION

But u need to PROVE it

We sure? Do you have a explanation/proof to that?

Ahh okay

I guess I have got one proof...

BY no solution, you mean no integral solution right ?

The proof that I have got is very simple

Shoot.

I will message u if u want

WOlfram tells me a solution, You must be meaning no integral solutions are there hmm ?

How did you enter that into wolfram? I couldnt get it to work.

http://www.wolframalpha.com/input/?i=%28n%5E%28n%2B1%29+-1%29%2F%28n-1%29+%3D+n%5En%5En

Sorry, I mean integer solution of n

hmm, that is what i assumed.

So yea we agreed there is no solution? as the only intersection is irrational.

Now, u just need to prove it

Surely this would suffice: n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0>n^n^n \[\forall n \le 1\] and n^n +n^(n-1) + n^(n-2)+...+n^2+n^1+n^0

ge is supposed to read as >=

But u directly said it without any proof

neatly done @sauravshakya