Prove that number is not
integer number .
A perfect square never ends in , therefore is not a perfect square, therefore is not an integer.
Come on Dhiab, that was too easy, where are the headcracking problems you used to post
PS : the proof for my original statement is quite trivial, a quick Google search will have it done.
Fine, I'll give it
Let be any integer number. is its square. Let us look at the last digit of , using arithmetic modulo . The following table enumerates all possible cases :
Sorry, I had to cut the table midway, it was too big for MHF.
This shows that a perfect square may only end by the digits and . If an integer does not terminate by neither of those digits, it cannot possibly be a perfect square.
This implies that if a number terminates by 3, it cannot be a perfect square.
I assume that the author forgot (or was unaware of) the Final Digit Rule for integer squares.Prove that number is not an integer.
If that is true, that number has a special rhythm to it: .
We have: .
And I suspect that we are expected to prove algebraically that is not a square.