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anonymous
 5 years ago
Someone who is good with proofs please talk to me
anonymous
 5 years ago
Someone who is good with proofs please talk to me

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am not good, but I will try the best I can.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok I will upload the problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01, 2, and 3 they are extra credit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Couldn't you just pick an a=9 and see if square root of 9^2 is the same as 9 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0R 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
 5 years ago
Best ResponseYou've already chosen the best response.0But if you are looking for book definition type proofs you got me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0well.. I think I got the solution of the first problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if it involves differentail equations, im outta here :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i thought your suppose to prove what he asks for

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0proof is a relative term.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0proof to a college teacher and proof to a 6 year old are not going to be the same.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think anny is doing a differential equation up there :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since 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
 5 years ago
Best ResponseYou've already chosen the best response.0you know the definition of the absolute value. right?.. does that make sense then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we can use what's this theorem we just proved to prove the next problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we 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
 5 years ago
Best ResponseYou've already chosen the best response.0I dunno if that makes any sense to you, but that's what I have anyway. I wish you get some points for that :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did you get the \[\sqrt{ab ^2{?}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, me too thank you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's the first theorem we proved, but instead of a, we have ab.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but how does it become ab^2 when its just IabI= IaI IbI

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From part one you showed that \[x = \sqrt{x^2}\] Let x = ab and you have it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the third one is easy.. I will give you a hint

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the following is known: \[a \le \left a \right, b \le \left b \right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just 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
 5 years ago
Best ResponseYou've already chosen the best response.0though it's easy to prove that using the definition of the absolute value.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what topic is this, thats how lost I am in my calculus class :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, if a \(\ge\)0 then what is a ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey lisaj you have to go back and see the definition of the absolute value.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0all proofs will make no sense to you if you are not aware of the definition.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@ polpak: IaI is idk i dont get it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if a is 5 what is lal?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the absolute value is 5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good. what if a=5, then its absolute value is?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So if a is less than 0, a = a. If a is greater than (or equal to) 0, a = a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you can see now that a is always greater than or equal to a?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so # 3 would be : ab\[ab \le \ \a b\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ab < or equal to IaIIbI

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x \ge 0 \implies x =x \implies x\ge x\] \[x \lt 0 \implies x = x \implies x \ge x\] \[\implies x \ge x \text{,}\forall x \in R\] Let ab be a product of real numbers \[\implies ab \in R \implies ab \ge ab \implies ?\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\left ab \right\ \ge \left a \right\left b \right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well sure, but a = a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is what i did the correct answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are trying to show that \(ab\le ab\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then it would be −a−b≤ab
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