{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Note we need a new q and a new r for this case now we

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e integer n, there are integers q and r, with 0 ≤ r < m so that n = qm + r. So m is fixed throughout the proof. We do an inducation on n. Base case: n = 0. In this case, n = 0m +0, so q = r = 0. Inductive step: Assume n = qm + r, with 0 ≤ r < m (inductive hypothesis). Prove: n+1 = q'm + r', with 0 ≤ r' < m. (Note: we need a new q and a new r for this case.) Now we start with n + 1 = qm + r + 1 from the inductive hypothesis. If 0 ≤ r +1 < m, then we can take r' = r+1 and q' = q. If r < m ­1, this is the case and we are done. In the remaining case of r = m ­1, then r+1 = m. So in this case n + 1 = qm + r + 1 = qm + m = (q+1)m + 0. So if we set q' = q+1 and r' = 0, we have proved what was needed in each case. QED. A very optional ad...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online