Prove P(n,m): m+n=n+m for Natural Numbers

[sli10126]

Homework Statement

Prove that P(n,m) m+n = n+m for all m,n in natural numbers.

Homework Equations

The Attempt at a Solution

I prove by induction.

Base case: P(0,0) = 0+0 = 0+0.
Inductive step: Let n be an arbitrary natural number. Suppose m+n =n+m. Adding 2 to both sides of the equation gives us m+n+2 = n+m+2.(end of proof)

My question is if this is sufficient enough as a proof. (The instructor hinted us to show P(0,0) first. Then show P(n,0) and then proceed to P(n,m). The hint confuses me.

 

[Mark44]

Homework Statement

Prove that P(n,m) m+n = n+m for all m,n in natural numbers.

Homework Equations

The Attempt at a Solution

I prove by induction.

Base case: P(0,0) = 0+0 = 0+0.
Inductive step: Let n be an arbitrary natural number. Suppose m+n =n+m. Adding 2 to both sides of the equation gives us m+n+2 = n+m+2.(end of proof)

My question is if this is sufficient enough as a proof. (The instructor hinted us to show P(0,0) first. Then show P(n,0) and then proceed to P(n,m). The hint confuses me.

You're OK with your base case, but you need to follow your instructor's suggestion.
Prove by induction on n that n + 0 = 0 + n; i.e., that the statement is true for P(n, 0).
Next, prove by induction on m that n + m = m + n.

 

[sli10126]

Would I need to show P(n+1,0) and P(0,m+1) or would P(n,0) and P(m) be sufficient? Because I know that for the inductive step we prove if P(n) then P(n+1).