Verified answer

You'll need to use the order axioms: (1) for x, y, z real if x < y and z > 0, then xz < yz; and (2) if x < y and y < z, then x < z.

Since a > 0 and 0 ≤ c ≤ d, the above axiom (1) tells us that a•0 = 0 ≤ ac ≤ ad. Similarly, as a < b and d ≥ 0, we have ad ≤ bd. Using axiom (2) above, we can conclude that

0 ≤ ac ≤ bd.

if 0 < a, 0 < c

so 0 < ac (product of 2 positives is positive)

if 0 < a and c = 0 then 0 = ac

hence if 0 < a < b and 0 ≤ c ≤ d. prove that 0 ≤ ac

a < b and if c = d then ac < bc or ac < db

if a < b and c < d then ac < bc and bc < bd so ac < bd

it extremely is finished by utilizing writing a evidence by utilizing induction. i've got not got time to place in writing up the formal evidence (it is what you will want for a discrete math college type, yet its an undemanding plug and chug format) For each thing else, you're able to do an casual evidence by utilizing induction. First you ought to do the backside case. you ought to instruct that at a=0 and/or c=0, the assertion is genuine. If a and c >= 0, then the two a and c must be effective or 0. If the two is 0, then ac must be 0. likewise, because of the fact b and d must be extra desirable than a and c, b and d must be extra desirable than 0. this ability that bd must be an outstanding selection and must be extra desirable than 0. Now you ought to instruct which you will amplify this good judgment for each possibility a,b,c and d. We do this by utilizing going up one step from 0, exhibiting that the assertion nevertheless works, and then exhibiting that it extremely works for each step. this is how we do it: notice that if x and y are the two effective numbers (i.e extra desirable than 0), then x+y>0. Likewise, x*y = x(a million)+x(2)+x(3)+x(4)+x(5)....x(y-2) + x(y-a million)+x(y) it is larger than 0. Now going returned on your difficulty, in case you be attentive to that b>a, then b+b must be extra desirable than a+a because of the fact the b's are increasing at a swifter value. Following this line of thinking, b+b+b is larger than a+a+a, b+b+b+b is larger than a+a+a+a. you could proceed this thinking till at last you get b+b+b+...(style of b's = d) is larger than a+a+a+...(style of a's = d). finally, because of the fact d>c, this ability that b+b+b+...(style of b's = c) is larger than a+a+a+...(style of a's =d), which could be rewritten as: ac<bd, winding up the evidence.

If 0 In C