I would love to see a proof for the division algorithm
@TheMathSorcerer
2 жыл бұрын
Great idea I should do that one!!!!
@whong09
2 жыл бұрын
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.
@GarryBurgess
2 жыл бұрын
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.
@TheMathSorcerer
2 жыл бұрын
A basic book on proof writing would do it. Most are pretty good. The book of proof is free and online😀
@iambic-kilometer
2 жыл бұрын
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.
@greense65
2 жыл бұрын
"... left to the reader". 🤣
@Satish-f3e
6 ай бұрын
sir 😊
@ReachByteBurst
2 жыл бұрын
wrong, 0 is neither
@proloycodes
2 жыл бұрын
wrong 0 is odd, because 0 = 2*0
@ReachByteBurst
2 жыл бұрын
@@proloycodes but aren't odd numbers defined as numbers that % 2 = 1?
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