Let x,y be elements of the Real Numbers such that x < y. There exists z in the Real Numbers such that x < z < y. Any thoughts on this proof?
Follow Math Help Forum on Facebook and Google+
Originally Posted by jstarks44444 Let x,y be elements of the Real Numbers such that x < y. There exists z in the Real Numbers such that x < z < y. Choose $\displaystyle z=(1/2)(x+y)$ and use the standard arithmetic and ordering axioms of $\displaystyle \mathbb{R}$ .
View Tag Cloud