To begin the proof, I first assume that f, and g both belong to the set of function F. This is because the proof asks me to prove the general statement for all f and g in F. Then I proceed by assuming the antecedent as usual. And from the antecedent, I know that f is lower-bounded by g, hence I can state its definition in the next line. Because I have to assume the antecedent and arrive at the consequent, I have to arrive at the definition that f^2 is lower-bounded by g^2. Hence I have to prove that
is true (where f(n) is f^2 and g(n) represents g^2). From this understanding, I have to pick values for both c and B. Since I don't really know how the proof will turn out, I left the pick c and pick B line empty for now and start by assuming the domain and the antecedent in the following line. Next, I squared f. Since from the antecedent, I know that f is greater than or equal to c'g, I also know that its square is greater than or equal to the square of c'g. In order to use the definition of f lower-bounded by g, n has to be greater than B', hence B can equal to B'. Now that I know what to pick for B, I can go back and write pick B= B'. And because B' is a natural number, B is also a natural number. Then I proceed with my previous calculation and got to that f^2 is greater or equal to c'^2 times g^2. In order to arrive at the consequent that f^2 is greater than or equal to cg^2, all I have to do is make c = c'^2. Hence I know what values to pick for c. And because c' is a positive real number, and real numbers are closed under multiplication, I can state that c is also a positive real number. Now that the proof is finished, all i have to do is to state the bookends and conclude the original claim.
In conclusion, it is much faster to figure out the values for c and B before tackling the problem. To be able to do this, quickly formulate the proof on a piece of scrap paper. Hence, by figuring out what I have to do in order to arrive at the consequent, I can write up the proof much faster than having to change my c and B values constantly.
is true (where f(n) is f^2 and g(n) represents g^2). From this understanding, I have to pick values for both c and B. Since I don't really know how the proof will turn out, I left the pick c and pick B line empty for now and start by assuming the domain and the antecedent in the following line. Next, I squared f. Since from the antecedent, I know that f is greater than or equal to c'g, I also know that its square is greater than or equal to the square of c'g. In order to use the definition of f lower-bounded by g, n has to be greater than B', hence B can equal to B'. Now that I know what to pick for B, I can go back and write pick B= B'. And because B' is a natural number, B is also a natural number. Then I proceed with my previous calculation and got to that f^2 is greater or equal to c'^2 times g^2. In order to arrive at the consequent that f^2 is greater than or equal to cg^2, all I have to do is make c = c'^2. Hence I know what values to pick for c. And because c' is a positive real number, and real numbers are closed under multiplication, I can state that c is also a positive real number. Now that the proof is finished, all i have to do is to state the bookends and conclude the original claim.
In conclusion, it is much faster to figure out the values for c and B before tackling the problem. To be able to do this, quickly formulate the proof on a piece of scrap paper. Hence, by figuring out what I have to do in order to arrive at the consequent, I can write up the proof much faster than having to change my c and B values constantly.

The entry of the slog was overall interesting and it was a good idea to post the hand-written work did by the student.
ReplyDeleteWow your slog is pretty good and also your hand writing is really good ~~
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