The equation
2t2+t3=t4
is satisfied by t=0
But if you cancel a t2 on both sides, making it
2+t=t2
t=0 is no longer a solution.
What gives? I thought nothing really changed, so the same solutions should apply.
Thanks
I have injection f:A→B and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...
No comments:
Post a Comment