16 prove the following statements for arbitrary x y z of an...

Name: 16 prove the following statements for arbitrary x y z of an ordered integral domain d a if x y and z 0 then xz yz b if x y 0 then x2 y2 14315
Uploaded: 2022-12-07T09:39:53-08:00
Duration: 15 min 35 s
Channel: Sri K
Description: 16 prove the following statements for arbitrary x y z of an ordered integral domain d a if x y and z 0 then xz yz b if x y 0 then x2 y2 14315

Transcript

00:01 To solve our first part, if x and y are non -negative, then we need to prove here, x plus y divided by a, i'm sorry, divided by 2 is equal to or greater than square root x .y.

00:28 This is the equation.

00:29 So now, as for the inequality, x plus y will be greater than or equal to 2 and of square root x y and this implies if we simplify this equation in the standard form minus 2 root xy which is greater than or equal to 0.

00:51 So as we know that if x and y are greater than and equal to 0, so first of all we will make this as algebra identity format.

01:02 So for that root x to the whole square will represent x.

01:06 Root y to the whole square will represent y and minus 2 root x root y.

01:13 So we are not changing our equation but we have placed this in the format of linear algebra identity.

01:21 And therefore we can close this algebra identity as root x minus root y to the whole square is greater than or equal to 0.

01:31 And we are using here a square minus b squared a square plus b square minus 2 a b which is equal to a minus b whole square this is our algebra identity and then we can state here as square of real number is positive therefore i am given value of root x minus root to y whole square is greater than or equal to z.

02:16 So now since x comma y are taken as non -negative, non -negative, so are value for root x and root y are real number.

02:38 So if they are real number, i will be for root x minus root y whole square will be positive.

02:50 And hence we will get the desire result which is x plus y divided by 2 is greater than not equal to root xy.

03:01 So hence we have proved our given equation.

03:08 This is our solution for 8 part of the question.

03:11 Now let's simplify our second part of the question which is in part b.

03:18 Part b we have here as given in the question, if a slash b and a slash b plus c then b is equal to a k1 and b plus c is equal to ak2 for some values of k1 k1 which are belonging to the z set for all natural numbers.

03:53 And b plus c minus b is equal to a k2 minus a k1.

04:00 And this implies is c is equal to a.

04:04 We will take here common k2.

04:08 B and b will be cancelled.

04:10 We are left with c only and here we will take common.

04:12 So this is k1.

04:14 And now 3c is equal to a.

04:20 We have multiplied 3, both the sides.

04:24 3k2, 3k1.

04:27 So since k1, k2 belongs to z set, then our value for 3, k2 minus k1 also belongs to z and hence we can say a slash 3c is true.

04:49 So this is a solution for b2.

04:52 Answer for b2.

04:54 Now let's simplify our next part which is part c.

04:58 So in the c part we have our value which is given over here as ab is greater than 0 and b c is less than 0.

05:11 So this implies is a is greater than 0, b is greater than 0 and at the same time b is greater than 0 and c is less than 0.

05:24 So now to show a x squared plus bx plus c is equal to 0 which is a quadratic equation and we know that it has two real parts.

05:43 So now when we factorize this equation as for completing this square method b by a x plus c by a we have divided the whole equation by a to make it x square as first term.

06:00 So now x squared plus b by 2a whole square plus c by a minus b square by 4 a square is equal to 0.

06:14 So now next value will be a plus b by 2a whole square plus 4ac minus b square divided by 4a square is equal to 0.

06:29 And since we know that a is greater than 0, b is greater than 0 and c is less than 0...

Question

(16) Prove the following statements for arbitrary x, y, z of an ordered integral domain D. (a) If $x > y$ and $z < 0$, then $xz < yz$. (b) If $x > y > 0$, then $x^2 > y^2$.

Question

(16) Prove the following statements for arbitrary x, y, z of an ordered integral domain D. (a) If $x > y$ and $z < 0$, then $xz < yz$. (b) If $x > y > 0$, then $x^2 > y^2$.

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