Define a set X of integers recursively as follows: Define a set X of integers recursively as follows:
Base: 5 is in X
Rule 1: If x is in X and x>0, then x+3 is in X
Rule2: If x is in X and x>0, then x+5 is in X
Show that every integer n>7 is in X
I am pretty new to this stuff and I am very lost on this. Please help!
 A: Let $n \ge 13$.  Divide $n$ by $5$ and take the remainder to get $n = k*5 + r$ and $k$ is at least $3$.
Case 0: if $r = 0$ then $n = 5+5+5+5 ..... = 5*k + 0 = n$ and $n $ is in $X$.
Case 3: if $r = 3$ then $n = 5+ 5+ 5+ .... + 3 = 5+ 3 + 5(k-1)$ is in $X$.
Case 1: if $r = 1$ then $n = 5+5+5+5+ ..... + 5 + 1 = 5+5+5+..... + 6 = 5(k-1)+3+3$ is in $X$
Can you do Case 2 and 4?
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This is a cute way.
Let $n = ......ba$ be the number written in digits.  $a$ is the last digt and $b$ is the second to last digit.
If $a = 0$ or $5$ then $n = 5+5+5+5+..... $ for some number of $5$ so $n $ is in $x$.
If $a = 1$ or $6$ then $n = 5+5+5+5+ .... + 5 + 3+3$ for some number of $5$ so $n$ is in $X$ if $n \ge 11$.
If $a = 2$ or $7$ then $n = 5+5+5+5+ ..... + 5 + 3+3+3+3$ for some number of $5$ so $n$ is in $n$ if $n > 12$.
If $a = 3$ or $8$ then $n = 5+5+5+... + 3$ and $n$ is in $X$ for $n \ge 8$.
If $a = 4$ or $9$ then $n = 5+5+5+5 + .... + 3+3+3$ and $n$ is in $X$ for $n \ge 14$.
So all integers greater than $12$ are in $X$.
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Ah, This is supposed to be a proof by induction!  (I was worried that might be advanced.)
1:  All numbers in the form $5k + 3j; k \ge 1; j \ge 0$ are in $X$.
Base case: It is true for $k = 1$ and $j=0$ as $5 = 5*1 + 0 \in X$.
Induction case: Assume it is true for $n = 5k + 3*0$.  Then by Rule one it is true for $n = 5k + 3*0 + 5 = 5(k+1) + 3*0$.  So it is true for all $n = 5k + 3*0$.
Assume it is true $n = 5k + 3j$.  Then by Rule two it is true for $n = 5k + 3j + 3 = 5k + 3(j+1)$ .  So it is true for all $n = 5k + 3j$.
2) Prove that for all $n\ge 13$ that $n = 5k + 3j$ for some $k \ge 1$ and $j \ge 0$ and either $k >1$ or $j > 2$.
Base case:  $13 = 5*2 + 3$
Induction:  If $n = 5k + 3j$ and $k > 1$ then$n+1 = 5(k-1) + 3(j+2)$ and $k \ge 1;j > 2$.  If $n = 5k + 3j$ and $j > 2$ then $n+1 = 5(k+2) + 3(j-3)$ and $k >1; j \ge 0$.
So All $n \ge 13$ are $n = 5k + 3j$ for some $k \ge 1$ and $j \ge 0$ and either $k > 1$ or $j > 2$.
A: It cannot be shown.
$X=\{5,8\}\cup\{n\in\mathbb Z\mid n\geq10\}$ satisfies base and rules but does not contain $9$.
