infinitely descending natural numbers Show that there is no infinitely descending sequence of natural numbers.
I was thinking that there exists no infinite descending chain on the natural numbers, since every chain of natural numbers has a minimal element. And so it reaches a finite minimum.
P.s I am just looking for a well built solution, since i cannot express the proof very clearly
 A: Suppose the first number in the sequence is $n$. Then there can be at most $n$ terms in the sequence, since the second term is at most $n-1$, the third $n-2$, etc.
A: The natural numbers are well-ordered by the usual $\le$ relation, that is, each subset $S\subseteq \mathbb{N}$ has a $\le$-least element. If there existed an infinitely descending sequence $a_1,a_2,\ldots$ of natural numbers, i.e., with $a_1>a_2>\cdots$, then $S=\{a_1,a_2,\ldots\}$ would not have a $\le$-least element. Contradiction.
A: What do you mean by natural number? Just the positive numbers? Or the positive numbers together with $0$?
Either way, you can prove it "by construction" or "by contradiction" (one of those, I'm not sure which). Choose any large natural number, say, $n = 10^{10^{10^{34}}}$ (that's a large number, can you imagine having that in dollars?) to be the first term of your infinitely descending sequence. The sequence will descend in steps of $1$. So the second term is $10^{10^{10^{34}}} - 1$. The  third term is $10^{10^{10^{34}}} - 2$. Then $10^{10^{10^{34}}} - 3$. You'd fall asleep at some point, but theoretically you'd eventually reach $10^{10^{10^{34}}} - 10^{10^{10^{34}}} + 1 = 1$. The next term would be $0$, which maybe you consider a natural number. But the next term would be $-1$, which you probably don't consider to be a natural number.
It's possible that you also consider the negative numbers to be natural numbers. If that's the case, I won't argue with you. The descent into negative numbers can be carried on indefinitely, but thankfully the thermometer can't go too far below $-40$.
A: A google search for
"induction and infinite descent"
results in many hits.
This one seems good
and is not behind a pay wall:
http://www.mathpages.com/home/kmath144/kmath144.htm
A: An integer is finite, and a finite number cannot infinitely decent, by definition
