"If a>b>0 (meaning that a>b and b>0) and , then ac > bd." (see ita-2-2-9-pg20)
"For a-b [is in R+] and c [is in R+], so ac-bc = (a-b)c [is in R+], and similarly c-d [is either in R+ or is zero] and b [is in R+] together imply that bc-bd [is either in R+ or is zero]; it necessarily follows that ac-bd = (ac-bc)+(bc-bd) [is in R+], that is ac>bd." (see ita-2-2-9-pg20)
Because: If bc-bd=0, then ac-bd = ac-bc which is in R+, and if bc-bd is in R+, then ac-bd is in R+ because both ac-bc and bc-bd are in R+.