For each n in the Naturals there exists m in the Naturals such that m > n. I'd really appreciate some help with the following proof. Thanks a bunch!
Follow Math Help Forum on Facebook and Google+
Originally Posted by jstarks44444 For each n in the Naturals there exists m in the Naturals such that m > n. I'd really appreciate some help with the following proof. Thanks a bunch! Simply take the succesor of n... Tonio
Can you please explain this? I don't know how to word what you are saying.... do I just introduce n+1?
Have you shown that $\displaystyle 0<1~?$ If so $\displaystyle n<n+1$. If not, what else can you use?
I'm not sure what you are implying here, I am fairly sure we can make the conclusion that 0 < 1, but what does that lead to?
Originally Posted by jstarks44444 I'm not sure what you are implying here, I am fairly sure we can make the conclusion that 0 < 1, but what does that lead to? Add $\displaystyle n$ to both sides to get $\displaystyle n<n+1$. Now $\displaystyle n+1$ is an integer greater than $\displaystyle n$.
View Tag Cloud