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

1. anonymous

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

2. anonymous

ok I will upload the problem

3. anonymous

4. anonymous

which problem?

5. anonymous

1, 2, and 3 they are extra credit

6. anonymous

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

7. 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.

8. anonymous

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

9. 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?

10. anonymous

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

11. anonymous

if it involves differentail equations, im outta here :)

12. anonymous

13. anonymous

proof is a relative term.

14. anonymous

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

15. anonymous

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

16. 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 $\sqrt {a^2}=\left| a \right|$

17. anonymous

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

18. anonymous

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

19. anonymous

oh ok

20. 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.

21. 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 :)

22. anonymous

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

23. anonymous

yes, me too thank you

24. anonymous

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

25. anonymous

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

26. anonymous

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

27. anonymous

28. anonymous

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

29. anonymous

good I wanna try

30. anonymous

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

31. 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.

32. anonymous

^^ I totally agree

33. anonymous

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

34. anonymous

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

35. anonymous

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

36. anonymous

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

37. anonymous

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

38. anonymous

@ polpak: IaI is idk i dont get it

39. anonymous

if a is 5 what is lal?

40. anonymous

the absolute value is 5

41. anonymous

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

42. anonymous

5

43. anonymous

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

44. anonymous

Right?

45. anonymous

true

46. anonymous

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

47. anonymous

so # 3 would be : ab$ab \le \ \-a |-b\|] 48. anonymous yes anny 49. anonymous ab < or equal to I-aII-bI 50. 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 $\implies ab \in R \implies |ab| \ge ab \implies ?$

51. anonymous

$\left| ab \right|\ \ge \left| -a \right|\left| -b \right|$

52. anonymous

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

53. anonymous

is what i did the correct answer?

54. anonymous

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

55. anonymous

so then it would be |−a||−b|≤ab