Prove that for all $n\ge9$, there exist natural numbers $x,y$ such that $n=2x+5y$.

Question

How would you use induction to prove this?

Answer 1

Hint: note that if $n\geq 9$ is written as $n=2x+5y$, then either $y\geq 1$ or $x\geq 2$. If the former, write $$ n+1=(2x+5y)+(2\cdot 3-5)=2(x+3)+5(y-1); $$ if the latter, write $$ n+1=(2x+5y)+(2\cdot(-2)+5)=2(x-2)+5(y+1). $$

Answer 2

Are you including zero as a natural number? I think it has to be to get 10

since 9 = 2 x 2 + 1 x 5

we can always make another 1 to add on with 5 - 2x2. If the number is even it can always be 2x, so to make the next odd number we can always take 2 from x and add set y = 1, for ever

Answer 3

If $n=9$ then $n = 2\cdot 2 + 5\cdot 1$; taking $x:= 2$ and $y:=1$ suffices. If $n \geq 9$ is an integer such that $n-1= 2x + 5y$ for some integers $x,y > 0$, then $n = n-1 + 1 = 2x+5y + 1 = 2x' + 5y'$. Note that $2x+5y+1 = 2x+5y+(5-4) = 2(x-2) + 5(y+1)$. So the preceding equalities are equivalent to $$ n = 2x' + 5y' = 2(x-2) + 5(y+1). $$ Taking $x' := x-2$ and $y' := y+1$ suffices, which are still integers.

Note that for $n=10$ we have $x' = 2-2 = 0$, which is not a natural number if you don't count $0$ as one such.