If a number x is even, then it is of the form 2n and so its square is 4n^2 and so can not be in the form 4y + 3.
If a number x is odd, then it is of the form 2n+1 and so its square is ...?
Can that form of x be of the form 4y + 3?
Prove the equation has no integer solutions.
So, I was thinking, prove by contradiction that statement is not true. So:
Assume has integer solutions.
This is where I get stuck. Do you let x and y be some integers k and j? How do you prove (or rather, disprove) integer solutions??
Assume there exist integers and that satisfy the equation.Prove the equation has no integer solutions.
We have: .
must either even or odd
If is even, then for some integer
. . Then: .
Hence, cannot be an integer.
If is odd, then , for some integer
. . Then: .
Hence, y cannot be an integer.
Therefore, no integer solution exists.
ok sorry - same part of homework: new question(s)
"If you get at least a C on the final exam, then you will pass the course." So if you got a D on the final - then what can you conclude?
He didn't pass OR there's still hope
OK: and second one, similar in nature:
"If you get at least a B on the final, then you will pass the course" You pass the course. What can you conclude?
He got at least a B on the final OR you can't conclude anything
(On this one, I'm inclined to think you can't conclude anything, because P does not cause Q, so he could still pass some other way.)