I'm in big trouble because of stupid math problen, help!!!
Harith Alanbari
problems which are due for very shortly and are graded. I've passed
hours on each one and still am stuck Here they are:
Ex 1
1) Prove that:
(a+b)/2 >= square root of (a*b)
2) Prove that (using questoin1) that:
(a+b)(b+c)(a+c) >= 8*a*b*c
3) Prove that:
if 1/h = 1/2 * (1/a +1/b) , then h =< square root of (a*b)
Ex 2
1) Prove that:
x*y =< (x*x + y*y) / 2
2)Prove that (using question 1):
x*y + y*z + z*x =< x*x + y*y +z*z
Thanks a million to anyone who can help, I really appreciate it, but I
just can't solve this problem...
Harlan Messinger
>I'm in serious trouble, because I can't solve any of these stupid
>problems which are due for very shortly and are graded. I've passed
>hours on each one and still am stuck Here they are:
If you can't do these problems, then why SHOULD you get as good a
grade as the people who can? If I hire what purport to be highly
marked graduates, and they turn out to have earned their marks by
getting other people to give them the answers, then how am I going to
get any decent work out of them? They'll just get fired--and then
having been fired will be on their employment record, which will make
it harder for them to get another job. It's in no one's best interest
to make people look like they know more than they do.
pengcong
(a+b)/2 * (a+b)/2 -a*b=(a*a2+2*a*b+b*b)/4-a*b=(a*a-2*a*b+b*b)/4=(a-b)/2 *
(a-b)/2>0so (a+b)/2 > square root of (a*b).
else so on.
Hauke Reddmann
<homework snipped>
There is no point of doing your assignment for
you, but I can give you some hints so you see
the _general approach_ to this problem class.
1. (x-y)*(x-y) >= 0 , because a square can't be negative.
Now we multiply out and get
2. x*x+y*y-2*x*y >= 0
Begins to resemble,eh?
So:
- If a square root haunts you, isolate it and square
both sides.
- It's a good idea to collect everything on one side
and look for factors then.
--
Hauke Reddmann <:-EX8 BRANDNEW,IMPROVED SIG!
Send all spam to bug...@thee.off
Send all personal e-mail to fc3...@math.uni-hamburg.de
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David Kastrup
> I'm in serious trouble, because I can't solve any of these stupid
> problems which are due for very shortly and are graded. I've passed
> hours on each one and still am stuck Here they are:
>
> Ex 1
> 1) Prove that:
> (a+b)/2 >= square root of (a*b)
Counterexample: a=b=-1.
> 2) Prove that (using questoin1) that:
> (a+b)(b+c)(a+c) >= 8*a*b*c
Counterexample: a=b=1, c=-1.
> 3) Prove that:
> if 1/h = 1/2 * (1/a +1/b) , then h =< square root of (a*b)
Plug in 1/a and 1/b instead of a and b in Ex 1).
Bonus question: why does the counterexample for 1) not work here?
--
David Kastrup Phone: +49-234-700-5570
Email: d...@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut für Neuroinformatik, Universitätsstr. 150, 44780 Bochum, Germany
David Kastrup
> Harith Alanbari <alan...@club-internet.fr> writes:
>
> > I'm in serious trouble, because I can't solve any of these stupid
> > problems which are due for very shortly and are graded. I've passed
> > hours on each one and still am stuck Here they are:
> >
> > Ex 1
> > 1) Prove that:
> > (a+b)/2 >= square root of (a*b)
>
> Counterexample: a=b=-1.
>
> > 2) Prove that (using questoin1) that:
> > (a+b)(b+c)(a+c) >= 8*a*b*c
>
> Counterexample: a=b=1, c=-1.
Of course, I meant a=1, b=c=-1.
Dennis Suchta
inequality
to prove the next two statements. Same for the second problem.
Harith Alanbari wrote:
> I'm in serious trouble, because I can't solve any of these stupid
> problems which are due for very shortly and are graded. I've passed
> hours on each one and still am stuck Here they are:
>
> Ex 1
> 1) Prove that:
> (a+b)/2 >= square root of (a*b)
> 2) Prove that (using questoin1) that:
> (a+b)(b+c)(a+c) >= 8*a*b*c
> 3) Prove that:
> if 1/h = 1/2 * (1/a +1/b) , then h =< square root of (a*b)
>
> Ex 2
> 1) Prove that:
> x*y =< (x*x + y*y) / 2
> 2)Prove that (using question 1):
> x*y + y*z + z*x =< x*x + y*y +z*z
>
Ingrid Voigt
> For (1) use the inequality and work back to a true statement.
This is not a proof method but handwaving. If you want to provea statement
you essentially have two choices: *Start* with a
true statements and work to your assertion. Or assume the
assertions is false and work to a false statement. What you
are suggesting may result in absolute nonsense.
Ingrid
ull...@math.okstate.edu
Actually assuming what you want to prove and working back
to something true is not a very reliable way to prove things. For
example it's easy to prove 1 = -1 this way:
Theorem 1 = -1
Proof:
1 = -1
1^2 = (-1)^2
1 = 1
QED.
The error in that proof is simply that showing that what you want
implies something true is not a valid way to prove things. (It
can of course be a useful way to _find_ the proof, but then you'd
better go back and write the proof forwards instead of backwards.)
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Mihir Wagle
1 follows from (\sqrt{a}-\sqrt{b})^2 >=0
2 follows from 1 logically
3 follows from 1 as in:
1/h = 1/2 (a+b)/ab >= 1/sqrt{ab}
Example 2 is the same effectively as example one... no assumption about
signs here though...
On Wed, 28 Oct 1998, Ingrid Voigt wrote:
> Dennis Suchta wrote:
>
> > For (1) use the inequality and work back to a true statement.
>
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Wilson Figueroa
>
> Harith Alanbari <alan...@club-internet.fr> writes:
>
> > I'm in serious trouble, because I can't solve any of these stupid
> > problems which are due for very shortly and are graded. I've passed
> > hours on each one and still am stuck Here they are:
> >
> > Ex 1
> > 1) Prove that:
> > (a+b)/2 >= square root of (a*b)
>
>
> > 2) Prove that (using questoin1) that:
> > (a+b)(b+c)(a+c) >= 8*a*b*c
>
>
> > 3) Prove that:
> > if 1/h = 1/2 * (1/a +1/b) , then h =< square root of (a*b)
>
>
> Bonus question: why does the counterexample for 1) not work here?
>
> David Kastrup Phone: +49-234-700-5570
> Email: d...@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
> Institut für Neuroinformatik, Universitätsstr. 150, 44780 Bochum, Germany
BRAVO!!!
Wonder what the instructor will thnik of this reply.
Perhaps the instructor doesn't understand the question!
Ronald Bruck
>
>BRAVO!!!
>
>Wonder what the instructor will thnik of this reply.
>
>Perhaps the instructor doesn't understand the question!
And perhaps the student who posted it didn't post the WHOLE assignment;
perhaps he didn't understand the importance of little words like
"positive".
Out of long experience, I can tell you that when students explain to me what
Professor X told them, it's often a very garbled account. It's always
best to consider the source.
BTW, I don't see how David Kastrup's counterexample a = b = 1, c = -1
contradicts (a+b)(a+c)(b+c) >= 8abc.
--Ron Bruck
Robert C. Cipriani
help. Did you never have to ask anyone for assistance? Highly unlikely...
Harlan Messinger wrote in message <36391742...@news.clark.net>...
>Harith Alanbari <alan...@club-internet.fr> wrote:
>
>>I'm in serious trouble, because I can't solve any of these stupid
>>problems which are due for very shortly and are graded. I've passed
>>hours on each one and still am stuck Here they are:
>
Dennis Suchta
statement to
a obviously "true" statement the steps can be reversed. Its the type of
arguement
almost as old as mathematics.
Ingrid Voigt wrote:
> Dennis Suchta wrote:
>
> > For (1) use the inequality and work back to a true statement.
>
> This is not a proof method but handwaving. If you want to provea statement
> you essentially have two choices: *Start* with a
> true statements and work to your assertion. Or assume the
> assertions is false and work to a false statement. What you
> are suggesting may result in absolute nonsense.
>
> Ingrid
>
> > Then use the
> > inequality
> > to prove the next two statements. Same for the second problem.
> >
> > Harith Alanbari wrote:
> >
> > > I'm in serious trouble, because I can't solve any of these stupid
> > > problems which are due for very shortly and are graded. I've passed
> > > hours on each one and still am stuck Here they are:
> > >
> > > Ex 1
> > > 1) Prove that:
> > > (a+b)/2 >= square root of (a*b)
> > > 2) Prove that (using questoin1) that:
> > > (a+b)(b+c)(a+c) >= 8*a*b*c
> > > 3) Prove that:
> > > if 1/h = 1/2 * (1/a +1/b) , then h =< square root of (a*b)
> > >
Dennis Suchta
steps
are not reversible. I could have written out proofs of the statements but
that is not
really the point. With what I wrote one could prove the statements and
should see
that they are related.
When presented with such statements one should always first test if they
are true
by substituing numbers and calculating (even if someone says "prove/show").
After gathering evidence, try to find EQUIVALENT statements by using
properties
of arithmetic that are "reversible".
ull...@math.okstate.edu wrote:
> In article <3637351B...@msn.com>,
> Dennis Suchta <d...@msn.com> wrote:
> > For (1) use the inequality and work back to a true statement.
>
Michael Press
>
> BTW, I don't see how David Kastrup's counterexample a = b = 1, c = -1
> contradicts (a+b)(a+c)(b+c) >= 8abc.
>
> --Ron Bruck
Neither did you trust that Mr. Kastrup is competent,
nor look a trifle further into the matter,
nor see that a = 1, b = c = -1 is a counter example.
--
Michael Press
pre...@apple.com
marvin carter
did someone else give you the solution
Dennis Suchta wrote:
> IF the basic properties of numbers/inequalities are used to reduce the
> statement to
> a obviously "true" statement the steps can be reversed. Its the type of
> arguement
> almost as old as mathematics.
>
> Ingrid Voigt wrote:
>
> > Dennis Suchta wrote:
> >
> > > For (1) use the inequality and work back to a true statement.
> >