For which natural numbers n do there exist nonnegative integers x and y such that n=4x+5y?
I used a proof by induction.
My proposition: For the natural numbers n>= 12 there exist nonnegative integers x and y such that n=4x+5y.
Initial step: Prove P(12) is true. Let x=3 and y=0 ... 12=4(3)+5(0), 12=12. So that checks out.
Now my trouble is on the inductive step.
Here is how my proof goes so far....
I had said that we will now prove that for n>=12 there exist nonnegative integers x and y such that n=4x+5y. We must first let P(n) be there exist x and y in the integers with x and y greater than or equal to zero, and x+y >= 4 such that n=4x+5y.
My first mistake: I used x = 3 and y = 0, which doesn't go with x+y>= 4, what can I do to fix that?
My proof continued: For inductive step, we prove that for all k in N with k>=12, if P(k) is true then P(K+1) is true. So let k be natural and assume P(k) or P(12) is true. I said we must show that K+1=4x+5y.
My second mistake: My instructor said "x and y are what" I'm assuming I can fix that by saying x and y are nonnegative integers?
My proof continued: Notice that 4(-1)+5=1. Since K=4x+5y...
My third mistake: My instructor said "same x and y??" What can I do to fix this or do differently.
My proof continued: K+1=4(x-1) + 5(y+1) by algebra. Since x and y are nonnegative integers, so are (x-1) and (y+1).
My final mistake, apparently: My instructor said "what if x = 0" So how can I fix this?
I will now finish the proof, to show you my work:
Thus, we have shown that K+1=4x+5y and therefore P(K+1) is true and inductive step has been established. Thus for the natural numbers n>=12, there exist nonnegative integers x and y with x+y>=4 such that n=4x+5y. Q.E.D.
We are allowed to submit corrections, so what errors are in my proof, can anyone help? it seems by my instructor's comments, and my work... I had almost got it?