Suppose m1 < m2 and n1 < n2. Show that m1n1 < m2n2.
ANy advice...THanks in advance.
Hello jzelltNot true. 3<8 and -2<-1, but -6>-8.
Perhaps you meant to say that these are natural numbers? In which case, $\displaystyle m_1 < m_2 \Rightarrow m_1 +m = m_2$, for some $\displaystyle m \in \mathbb{N}$, and $\displaystyle n_1 < n_2 \Rightarrow n_1+n = n_2,$ for some $\displaystyle n \in \mathbb{N}$.
So $\displaystyle m_2n_2 = (m_1 +m)(n_1 + n)$
$\displaystyle = m_1n_1 + mn_1+nm_1+mn$
$\displaystyle > m_1n_1$
Grandad
A bit simpler, I think: If $\displaystyle m_1< m_2$, and $\displaystyle n_1> 0$ then $\displaystyle m_1n_1< m_2n_1$. If $\displaystyle n_1< n_2$ and $\displaystyle m_2> 0$ then $\displaystyle m_2n_1< m_2n_2$. Those together ("<" is a transitive relation) give $\displaystyle m_1n_1< m_2n_2$.
Notice that does not say anything about $\displaystyle m_1$. This only requires that 3 of the four numbers be positive.