Question

Someone who is good with proofs please talk to me

Answer 1

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

Answer 2

ok I will upload the problem

Answer 3

which problem?

Answer 4

1, 2, and 3 they are extra credit

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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?

Answer 9

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

Answer 10

if it involves differentail equations, im outta here :)

Answer 11

i thought your suppose to prove what he asks for

Answer 12

proof is a relative term.

Answer 13

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

Answer 14

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

Answer 15

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
\[\sqrt {a^2}=\left| a \right|\]

Answer 16

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

Answer 17

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

Answer 18

oh ok

Answer 19

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.

Answer 20

I dunno if that makes any sense to you, but that's what I have anyway. I wish you get some points for that :)

Answer 21

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

Answer 22

yes, me too thank you

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

what about # 3

Answer 27

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

Answer 28

good I wanna try

Answer 29

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

Answer 30

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.

Answer 31

^^ I totally agree

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

@ polpak: IaI is idk i dont get it

Answer 38

if a is 5 what is lal?

Answer 39

the absolute value is 5

Answer 40

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

Answer 41

5

Answer 42

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

Answer 43

Right?

Answer 44

true

Answer 45

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

Answer 46

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

Answer 47

yes anny

Answer 48

ab < or equal to I-aII-bI

Answer 49

\[x \ge 0 \implies |x| =x \implies |x|\ge x\]
\[x \lt 0 \implies |x| = -x \implies |x| \ge x\]
\[\implies |x| \ge x \text{,}\forall x \in R\]
Let ab be a product of real numbers
\[\implies ab \in R \implies |ab| \ge ab \implies ?\]

Answer 50

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

Answer 51

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

Answer 52

is what i did the correct answer?

Answer 53

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

Answer 54

so then it would be |−a||−b|≤ab