Prove For all k m and n in N if m n then mk nk and mk nkS

Prove: For all k, m, and n in N, if m < n, then m+k < n+k and mk < nk.

Solution

Let, m+k>=n+k

Adding -k to both sides gives

m+k+(-k)>=n+k+(-k)

m+(k-k)>=n+(k-k)

m+0>=n+0

m>=n

which is a contradiction.

Hence, m+k<n+k

m<n

m-n<0

For all k in N we have k >0 hence

(m-n)k<0

mk-nk<0

mk<nk

Hence proved