Wikipedia:Reference desk/Archives/Mathematics/2022 October 3

= October 3 =

True or not?
Is it true that if (a/b) = (c/d) where a>c and b>d that ab/cd = a2? I have tried with lot of examples and it came true each time. Exclusive Editor  Notify Me! 09:01, 3 October 2022 (UTC)


 * Try a=b=3, c=d=2. Then ab/(cd) = 9/4, which is not a2. Staecker (talk) 10:33, 3 October 2022 (UTC)
 * More generally: $$a = \frac{bc}{d} \Longrightarrow a^2 = \frac{abc}{d}$$, so c is in the numerator, not the denominator. The inequalities do not enter at all. --Wrongfilter (talk) 11:57, 3 October 2022 (UTC)
 * So equality holds if and only if $$|c|=1.$$ For example, when $$a=3,b=21,c=1,d=7,$$ we have
 * $$\frac{a}{b}=\frac{3}{21}=\frac{3\times 1}{3\times 7}=\frac{1}{7}=\frac{c}{d}.$$
 * Then
 * $$\frac{ab}{cd}=\frac{3\times 21}{1\times 7}=\frac{63}{7}=\frac{7\times 9}{7\times 1}=\frac{9}{1}=9=3^2=a^2.$$
 * --Lambiam 13:39, 3 October 2022 (UTC)