Originally Posted by

**JBrandon** so the full proof should be.

Let $\displaystyle A \neq \phi$, $\displaystyle L=\text{LUB}(A)$, $\displaystyle a \in A$, $\displaystyle pa \in pA$, and p be a positive number.

Since$\displaystyle A \neq \phi$, and $\displaystyle L=\text{LUB}(A)$, then $\displaystyle a \le L$.

Since $\displaystyle a \le L$, then $\displaystyle pa \le pL$.

therefor $\displaystyle pL=\text{UB}(pA)$.

Assume $\displaystyle \lambda=\text{LUB}(pA)$

since $\displaystyle \lambda=\text{LUB}(pA)$, and $\displaystyle pL=\text{UB}(pA)$, then $\displaystyle \lambda \le pL$.

since $\displaystyle \lambda \le pL$, then $\displaystyle \frac{\lambda}{p} < L$

since $\displaystyle \frac{\lambda}{p} < L$, and $\displaystyle L=\text{LUB}(A)$, then $\displaystyle \frac{\lambda}{p} < a \le L$

since $\displaystyle \frac{\lambda}{p} < a \le L$, then $\displaystyle \lambda < pa \le pL$. a contradiction as $\displaystyle pa \le \lambda$ must be true for $\displaystyle \lambda=\text{LUB}(pA)$

therefor $\displaystyle pL < \lambda$ and $\displaystyle pL = \text{LUB}(pA)$