f:R-->R and f '(x)<= a < 0 for any x

show that f is onto R

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Proof: since f '(x) < 0, we have f(x) a monotone decreasing function

since f '(x) exists, we know f(x) must be continuous for any x in R

the cardinality of two like sets is the same so: |R| = |R|

Since the cardinality of the domain and range is the same, proving onto amounts to proving that f:R-->R is one-to-one

But f(a)>f(b) for any a,b in R

Thus, f:a-->f(a) and f:b-->f(b), and

f(a) is not equal to f(b) whenever a is not equal to b

which is to say f(x) is one-to-one for any x in R

Since the cardinality of the domain equals the cardinality of the range (i.e. |R|=|R|) and f is one-to-one,

f(x) must be ONTO.

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Does anyone know what is wrong with this proof if anything? Thank you