anonymous
  • anonymous
Someone who is good with proofs please talk to me
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
I am not good, but I will try the best I can.
anonymous
  • anonymous
ok I will upload the problem
anonymous
  • anonymous
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anonymous
  • anonymous
which problem?
anonymous
  • anonymous
1, 2, and 3 they are extra credit
anonymous
  • anonymous
Couldn't you just pick an a=-9 and see if square root of -9^2 is the same as |-9| ?
anonymous
  • anonymous
R means any number, so try to pick numbers that you think would make it bad, and if they are good..use that as proof.
anonymous
  • anonymous
But if you are looking for book definition type proofs you got me.
anonymous
  • anonymous
I think that when it says show that it wants you to pick numbers from the real set and try to make it false. Is that your understanding too?
anonymous
  • anonymous
well.. I think I got the solution of the first problem
anonymous
  • anonymous
if it involves differentail equations, im outta here :)
anonymous
  • anonymous
i thought your suppose to prove what he asks for
anonymous
  • anonymous
proof is a relative term.
anonymous
  • anonymous
proof to a college teacher and proof to a 6 year old are not going to be the same.
anonymous
  • anonymous
I think anny is doing a differential equation up there :)
anonymous
  • anonymous
since a^2=(+a)^2=(-a)^2, the numbers a, -a are square roots of a^2.. that's: \[a \ge0 \implies \sqrt{ a^2}=a\] \[a<0 \implies \sqrt {a^2}=-a\] which is the same as the definition of the absolute value, so \[\sqrt {a^2}=\left| a \right|\]
anonymous
  • anonymous
you know the definition of the absolute value. right?.. does that make sense then?
anonymous
  • anonymous
we can use what's this theorem we just proved to prove the next problem.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
we want to prove that: \[\left| ab \right|=\left| a \right|.\left| b \right|\] but we just proved that: \[\left| ab \right|=\sqrt {(ab)^2}\] we can now proceed, \[\sqrt {(ab)^2}=\sqrt {a^2b^2}=\sqrt {a^2} . \sqrt{b^2}=\left| a \right|.\left| b \right|\] that's it.
anonymous
  • anonymous
I dunno if that makes any sense to you, but that's what I have anyway. I wish you get some points for that :)
anonymous
  • anonymous
where did you get the \[\sqrt{ab ^2{?}}\]
anonymous
  • anonymous
yes, me too thank you
anonymous
  • anonymous
it's the first theorem we proved, but instead of a, we have ab.
anonymous
  • anonymous
but how does it become ab^2 when its just IabI= IaI IbI
anonymous
  • anonymous
From part one you showed that \[|x| = \sqrt{x^2}\] Let x = ab and you have it.
anonymous
  • anonymous
what about # 3
anonymous
  • anonymous
the third one is easy.. I will give you a hint
anonymous
  • anonymous
good I wanna try
anonymous
  • anonymous
the following is known: \[a \le \left| a \right|, b \le \left| b \right|\]
anonymous
  • anonymous
Just have to go through the combinations\(a\ge 0, a \lt 0, b \ge 0, b \lt 0\) I'm not sure you can assume that \(x \le |x|\) otherwise you'd just let x = ab and you've already got it. I think you have to show that \(x \le |x|\) using the definition of the absolute value.
anonymous
  • anonymous
^^ I totally agree
anonymous
  • anonymous
though it's easy to prove that using the definition of the absolute value.
anonymous
  • anonymous
what topic is this, thats how lost I am in my calculus class :(
anonymous
  • anonymous
Ok, if a \(\ge\)0 then what is |a| ?
anonymous
  • anonymous
hey lisaj you have to go back and see the definition of the absolute value.
anonymous
  • anonymous
all proofs will make no sense to you if you are not aware of the definition.
anonymous
  • anonymous
@ polpak: IaI is idk i dont get it
anonymous
  • anonymous
if a is 5 what is lal?
anonymous
  • anonymous
the absolute value is 5
anonymous
  • anonymous
good. what if a=-5, then its absolute value is?
anonymous
  • anonymous
5
anonymous
  • anonymous
So if a is less than 0, |a| = -a. If a is greater than (or equal to) 0, |a| = a
anonymous
  • anonymous
Right?
anonymous
  • anonymous
true
anonymous
  • anonymous
so you can see now that |a| is always greater than or equal to a?
anonymous
  • anonymous
so # 3 would be : ab\[ab \le \ \-a |-b\|]
anonymous
  • anonymous
yes anny
anonymous
  • anonymous
ab < or equal to I-aII-bI
anonymous
  • anonymous
\[x \ge 0 \implies |x| =x \implies |x|\ge x\] \[x \lt 0 \implies |x| = -x \implies |x| \ge x\] \[\implies |x| \ge x \text{,}\forall x \in R\] Let ab be a product of real numbers \[\implies ab \in R \implies |ab| \ge ab \implies ?\]
anonymous
  • anonymous
\[\left| ab \right|\ \ge \left| -a \right|\left| -b \right|\]
anonymous
  • anonymous
Well sure, but |a| = |-a|
anonymous
  • anonymous
is what i did the correct answer?
anonymous
  • anonymous
You are trying to show that \(ab\le |a||b|\)
anonymous
  • anonymous
so then it would be |−a||−b|≤ab

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