Someone who is good with proofs please talk to me

- anonymous

Someone who is good with proofs please talk to me

- schrodinger

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- anonymous

I am not good, but I will try the best I can.

- anonymous

ok I will upload the problem

- anonymous

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## More answers

- anonymous

which problem?

- anonymous

1, 2, and 3 they are extra credit

- anonymous

Couldn't you just pick an a=-9 and see if square root of -9^2 is the same as |-9| ?

- 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

But if you are looking for book definition type proofs you got me.

- 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

well.. I think I got the solution of the first problem

- anonymous

if it involves differentail equations, im outta here :)

- anonymous

i thought your suppose to prove what he asks for

- anonymous

proof is a relative term.

- anonymous

proof to a college teacher and proof to a 6 year old are not going to be the same.

- anonymous

I think anny is doing a differential equation up there :)

- 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

you know the definition of the absolute value. right?.. does that make sense then?

- anonymous

we can use what's this theorem we just proved to prove the next problem.

- anonymous

oh ok

- 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

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

where did you get the \[\sqrt{ab ^2{?}}\]

- anonymous

yes, me too thank you

- anonymous

it's the first theorem we proved, but instead of a, we have ab.

- anonymous

but how does it become ab^2 when its just IabI= IaI IbI

- anonymous

From part one you showed that \[|x| = \sqrt{x^2}\]
Let x = ab and you have it.

- anonymous

what about # 3

- anonymous

the third one is easy.. I will give you a hint

- anonymous

good I wanna try

- anonymous

the following is known:
\[a \le \left| a \right|, b \le \left| b \right|\]

- 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

^^ I totally agree

- anonymous

though it's easy to prove that using the definition of the absolute value.

- anonymous

what topic is this, thats how lost I am in my calculus class :(

- anonymous

Ok, if a \(\ge\)0 then what is |a| ?

- anonymous

hey lisaj you have to go back and see the definition of the absolute value.

- anonymous

all proofs will make no sense to you if you are not aware of the definition.

- anonymous

@ polpak: IaI is idk i dont get it

- anonymous

if a is 5 what is lal?

- anonymous

the absolute value is 5

- anonymous

good. what if a=-5, then its absolute value is?

- anonymous

5

- anonymous

So if a is less than 0, |a| = -a. If a is greater than (or equal to) 0,
|a| = a

- anonymous

Right?

- anonymous

true

- anonymous

so you can see now that |a| is always greater than or equal to a?

- anonymous

so # 3 would be : ab\[ab \le \ \-a |-b\|]

- anonymous

yes anny

- anonymous

ab < or equal to I-aII-bI

- 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

\[\left| ab \right|\ \ge \left| -a \right|\left| -b \right|\]

- anonymous

Well sure, but |a| = |-a|

- anonymous

is what i did the correct answer?

- anonymous

You are trying to show that \(ab\le |a||b|\)

- anonymous

so then it would be |−a||−b|≤ab

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