Question

how to prove if x>=0 and x<= c for all c>0, then x=0

Answer 1

you can prove it by contradiction, suppose that
x>=0 and x<= c for all c>0, *and* x is not equal to zero. so there are two cases, x > 0 (you get contradiction) and x < 0 , you get contradiction

Answer 2

couldn't be the case that x is just equal to c ?

Answer 3

one sec

Answer 4

no it couldnt because it says for all c > 0

Answer 5

so basically what contradicts in the case of x>0 is that c is not a certain number whereas x is a certain number?

Answer 6

prove if x>=0 and x<= c for all c>0, then x=0
proof (by contradiction): suppose that x>=0, x<= c for all c>0, *and* x is not equal to zero
now if x is not equal to zero then you have two cases. either x > 0 or x < 0 . suppose that x > 0, then we have
0 < = x <= c , for all c > 0 .
Now let choose a c, c = x - e that is still greater than zero.
now
x-e < x , but now c < x and x <= c from above. contradiction

Answer 7

each x is chosen and fixed, once you pick x. but c is variable, since it says for all c > 0

Answer 8

it might be easier if i show it on a number line , the case x > 0

Answer 9

i understand that. i just dont know how to put it in words. lol