Question

Please, help
if a >b
Prove \(a^2b 11 years ago

Answer 1

Alrighty then.

Answer 2

@xapproachesinfinity help me, friend?

Answer 3

so far, I got
a > b
a^3 > b^3 --> a^3 -b^3 >0 --> \(\dfrac{a^3-b^3}{3}> 0\)
and \(\dfrac{a^3-b^3}{3}+ab^2 -ab^2> 0\\\dfrac{a^3-b^3}{3}+ab^2>ab^2\)

Answer 4

but ab^2 is not > a^2b , so that I am stuck

Answer 5

hmm try to show that ab^2>a^2b

Answer 6

\(\large a>b\)

\(\large b<a\)

\(\large 0<a-b\)

\(\large 0<(a-b)^2\)

\(\large 0<(a^2-2ab+b^2)\)

\(\large 0<(a^2-3ab+ab+b^2)\)

\(\large 3ab<(a^2+ab+b^2)\)

\(\large 3ab(a-b)<(a-b)(a^2+ab+b^2)\)

\(\large 3ab(a-b)<(a^3-b^3)\)

\(\large (a^2b-ab^2)<(\dfrac{a^3-b^3)}{3}\)

Answer 7

hahaha... how??
a > b
a^2 > ab
a^2b > ab^2
not conversely

Answer 8

eh he got the answer lol

Answer 9

cool

Answer 10

Yes, it 's perfect. Thank you

Answer 11

yw

Answer 12

I'm sure you go your way too

Answer 13

can*

Answer 14

what you said here is not true
a>b then a^2>ab
we have no idea what a is positive or negative

Answer 15

yes, you are right.

Answer 16

eh it is really hard to go your way! not knowing about a and b except that a>b

Answer 17

Hey, help me make the next problem clear, ok?

Answer 18

you got another problem hehe

Answer 19

I confused!!!
the problem: Prove that in an ordered field , if \(\sqrt 2\) is a positive number whose square is 2, then \(\sqrt 2 <3/2\)

Answer 20

I don't see any mistake on my logic, but it leads me to nowhere

Answer 21

Suppose \(\sqrt 2 > 3/2\) then \(\sqrt 2 - 3/2 >0\)
square a positive number gives us a positive number also. So that
\((\sqrt 2 -3/2)^2>0\)
it gives me \(2 -3\sqrt 2+9/4 >0\)
that means \(17/4 > 3\sqrt 2\) or \(17/12 > \sqrt 2\)
since both sides are >0, we square it , the sign doesn't change and I get 2.0006 > 2
:((((

Answer 22

I'm really sleepy.

Answer 23

ok, go to bed, friend. :)

Answer 24

you started with \(\sqrt{2}>3/2 \) then you reached \(\sqrt{2}<17/12<3/2\)
so contradiction here

Answer 25

the problem lies in that step!

Answer 26

you can stop in there and go back the conclusion that \(\sqrt{2}>3/2\) is not true
\(\therefore \sqrt{2}<3/2\)

Answer 27

Gotttttttttttt you, haaahaa... I didn't apply brake on my run

Answer 28

yeah you were running fast down hill hehehe

Answer 29

number theory?

Answer 30

advance cal . prove everything :)

Answer 31

hehe! your course is dealing with pure proofs hehe

Answer 32

proves*

Answer 33

this is the general,lol
\(\large \sqrt{2}<\dfrac{3}{2}\)

\(\large \ 2 \times \sqrt{2}<3\)

\(\large \sqrt{2 \times 4}<3\)

\(\large \sqrt{8}<\sqrt{9}\)

\(\large \ 8<9\)

Answer 34

cant understand that ordered field

Answer 35

hehe actually you next step is fine! since 2<2.0006 is not true
therefore the assumption is not true

Answer 36

@xapproachesinfinity 2 < 2.0006 is true

Answer 37

@mathmath333 ordered field means <

Answer 38

eh my bad not paying the attention lol

Answer 39

It 's a field

Answer 40

exhausted hehe

Answer 41

so that it follows some rule.

Answer 42

you can't really go conversely! can you?

Answer 43

@xapproachesinfinity go to bed, friend. hahahaha

Answer 44

lol

Answer 45

I HAVE to use the rule of if a <b then a -b <0,

Answer 46

and manipulate everything base on it.

Answer 47

for example, if a^2 -b^2 <0, I must break it as (a+b)(a-b)<0
it <0 when one of them >0 and other <0 , so on and so on

Answer 48

eh i should get some rest haha!

Answer 49

sleeeeeeeep !! you are lucky. I have a bunch of homework due on Monday. I have to work out all of them.

Answer 50

i have a bunch too! but easy stuff you know hehe

Answer 51

:)

Answer 52

We all should sleep.
I'll bring the drugged popcorn.