Since I had no number theory, I had that feeling of not understanding what was being presented to me. That means, I would have great difficulty creating any meaningful question to lead me to more understanding. There must be some book somewhere to take away this feeling.
Observation: 0 = 2(0) is even and 1 = 2(0)+1 is odd. Sketch of proof not requiring the division algorithm. Suppose positive counterexamples exist and let s be the least positive counterexample to the result; here, we are using the principle that every nonempty set of positive integers has a least element. We note that 1 is odd, so it is not a counterexample. Hence, s>=2. Also, s-1>=1 is not a counterexample because it is positive and below s, so s-1 is either even or odd. If s-1 is even, i.e., s-1=2k, then s=2k+1 is odd which is a contradiction. If s-1 is odd, i.e. s-1=2k+1, then s=2k+2=2(k+1) is even which is also a contradiction. Each case leads to a contradiction, so no positive counterexample exists. The proof that no negative counterexample exists is similar and left to the reader.
I would love to see a proof for the division algorithm
Great idea I should do that one!!!!
Since I had no number theory, I had that feeling of not understanding what was being presented to me. That means, I would have great difficulty creating any meaningful question to lead me to more understanding. There must be some book somewhere to take away this feeling.
A basic book on proof writing would do it. Most are pretty good. The book of proof is free and online😀
Observation: 0 = 2(0) is even and 1 = 2(0)+1 is odd.
Sketch of proof not requiring the division algorithm.
Suppose positive counterexamples exist and let s be the least positive counterexample to the result; here, we are using the principle that every nonempty set of positive integers has a least element. We note that 1 is odd, so it is not a counterexample. Hence, s>=2. Also, s-1>=1 is not a counterexample because it is positive and below s, so s-1 is either even or odd. If s-1 is even, i.e., s-1=2k, then s=2k+1 is odd which is a contradiction. If s-1 is odd, i.e. s-1=2k+1, then s=2k+2=2(k+1) is even which is also a contradiction. Each case leads to a contradiction, so no positive counterexample exists. The proof that no negative counterexample exists is similar and left to the reader.
"... left to the reader". 🤣
If x % 2 is 1 then (x + 1) % 2 is 0. If x % 2 is 0 then (x + 1) % 2 is 1. Integers are closed under addition by 1, 0 % 2 is 0. Three dots.
sir 😊
wrong, 0 is neither
wrong 0 is odd, because 0 = 2*0
@@mojacodes but aren't odd numbers defined as numbers that % 2 = 1?
@@mojacodes so it actually should be even
@@ReachByteBurst oh yeah my mistake
@@mojacodes aight