a) f = {(n,m) element of N x N | m is the integer obtained from n by reversing i

Question

a) f = {(n,m) element of N x N | m is the integer obtained from n by reversing its digits}, e.g. f (123) = 321 and f (11271) = 17211

b) f = {(a/b, 2^a3^b) element of Q^+ x N}, where Q^+ = {q element of Q | q > 0}

Explanation / Answer

if you want to prove something not to be the case, you only need to show one counterexample, i.e. you do not have to disprove the statement for all possible values of a and b. Let f(x) = x^2 ==> f(a+b) = (a+b)^2... It is crucial that you understand this. Just replace the x's (in this case just one x) in the formula for the function, i.e. f(x) = x^2 with the specified x-value, in this case a+b. f(a+b) = (a+b)^2 = a^2 +2ab + b^2 = a^2 + b^2 + 2ab f(a) = a^2 f(b) = b^2 ==> f(a) + f(b) = a^2 + b^2 As you can see f(a+b) ? f(a) + f(b). Now see if you can find another function to prove that f(a+b) ? f(a) + f(b). You can, however, find a function, e.g. f(x) = x that will show that in this particular case f(a+b) = f(a) + f(b)....see that this is so for yourself. But this is immaterial because of the counterexample. Can you find other functions such that f(a+b) = f(a) + f(b)

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