Question

F1 is a set of functions of one real variable defined recursively as follows. • Base case of(x) = 1, then f € F1 o f(x)= x, then f € F1 . Constructor case of, ge F1, then fx ge F1 and f + g € F1 a. Prove by ordinary induction that f(x) = nx +n, n >= 1 is in F1 b. Prove by structural induction that if f(x) is in F1, then f(0) >= 0 .​

Answer 1

Answer

\begin{gathered} \mathrm {8101 = is \: not \: multipe \: of \: 9 } \\\end{gathered}8101=isnotmultipeof9

\begin{gathered} \mathrm {1246 = is \: not \: multipe \: of \: 9 } \\\end{gathered}1246=isnotmultipeof9

\begin{gathered} \mathrm {4546 = is \: not \: multipe \: of \: 9 } \\\end{gathered}4546=isnotmultipeof9

\begin{gathered} \mathrm {1008 = is \: \: multipe \: of \: 9 } \\\end{gathered}1008=ismultipeof9

\mathrm \red{Formula}Formula

Sum Of All Numbers

If The Total of Numbers is multiple of table 9 means it is divisible with 9

\mathrm \green{Proof}Proof

(a) 8101 ➢ 8 + 1 + 0 + 1 = 10

( 10 is not multiple of 9, so 8101 is not multiple of 9 )

(b) 1246 ➢ 1 + 2 + 4 + 6 = 13

( 13 is not multiple of 9, so 1246 is not multiple of 9 )

(c) 4546 ➢ 4 + 5 + 4 + 6 = 19

( 19 is not multiple of 9, so 4546 is not multiple of 9 )

(d) 1008 ➢ 1 + 0 + 0 + 8 = 9

( 9 is multiple of 9, so 1008 is multiple

Answer:

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Answer 2

A Recursive Function is defined as: We say the function f is defined recursively if the value of f at 1 is specified and if a rule is provided for determining f(n+1) from f(n).

Explanation:

The question implies a 'general' proof that can be used to prove both parts. Please find it as follows:

                 $Let \ f:N \to R\\\\f': N \to R\\\\g: R \to R\\\\f(1) = f'(1) = x_1 \in R\\\\f(x+1) = g(f(x))\\\\f(x+1) = g(f'(x))\\\\Whereas\\\\f(1) = x_1; f(2) = g(x_1); f(3) = g(g(x_1)); f(4) = g(g(g(x_1)))\\\\similarly \ for\ f'\\\\Prove \ by \ usual \ induction \\that \ for \ all \ x \in N\\we \ shall \ have f(x) = f'(x)\\\\Base \ case \ can \ be \ proven \ by \\\f(1) = f'(1)\\\\Inductive \ step:\\\\Suppose \ f(k) = f'(k)\\use \ the \ supposition \ to \ prove\\f(k+1) = f'(k+1)$