Stupid question (1 Viewer)

C

coca cola

Guest
How do you prove (x)^(4/3) >= 0, for all real x.

Is it because [(x)^(1/3)]^4 >= 0

and since (x)^(1/3) is a vaild operation for all real x.

But if the denominator is even instead of being odd, then this wouldn't work since x is gotta be bigger or equal to zero.

Am I right? Thanks.
 

Archman

Member
Joined
Jul 29, 2003
Messages
337
Gender
Undisclosed
HSC
N/A
u can never use induction to prove "for all real x"
 
C

coca cola

Guest
Induction is only for positive intergers, but if its a induction proof we wouldn't be proving for x right? We'll be proving for 2n+1(odd) for the denominator or something. Tis not a proof question anyways, I was just wondering if I got the logics right.
 

mathock

Member
Joined
Oct 21, 2004
Messages
142
Gender
Male
HSC
2004
haha by logic, x can only equal zero... i dont understand wats being asked :(
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
coca cola said:
Induction is only for positive intergers, but if its a induction proof we wouldn't be proving for x right? We'll be proving for 2n+1(odd) for the denominator or something. Tis not a proof question anyways, I was just wondering if I got the logics right.
Induction can be used to prove negative integers.
It should be able to be used to prove non-integers as well, but not all real number.
You can only prove numbers that go with a fixed increment.
For example, prove for n=k and k=k+0.5.
If you can also establish the truth for n=pi,
its true for n=pi, n=pi+0.5, n=pi+1, n=pi+1.5, etc
 
S

Shuter

Guest
mojako said:
Induction can be used to prove negative integers.
Yeah I was thinking you could use induction to prove for >0 then to prove for <0. Hmm but yeah i wasn't thinking and to prove for all real X you'd need the increment to be lim->0.

Disreguard my thoughts.
 

gordo

Resident Jew
Joined
Feb 5, 2004
Messages
2,352
Location
bondi, sydney
Gender
Male
HSC
2004
since it is the cube root of x to the 4, if x is neagtive the power of 4 will make is positive, if x is positive it will staty positive, therefore it will be greater or = to zero
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
coca cola said:
But if the denominator is even instead of being odd, then this wouldn't work since x is gotta be bigger or equal to zero.
If an expression is written as sqrt(x^2), it must be >=0 and is identical to |x|
Simlarly I would expect x^(2/2) to be >=0
and x^(a/b), where both a and b are even numbers, to be >=0
 
C

coca cola

Guest
mojako said:
If an expression is written as sqrt(x^2), it must be >=0 and is identical to |x|
Simlarly I would expect x^(2/2) to be >=0
and x^(a/b), where both a and b are even numbers, to be >=0
umm I don't see your point. sqrt(anything) can be negative or positive. Even if that something is a square. i.e. sqrt(4^2) = + or - 4.

note: |x| = postive root of (x), not sqrt(x).

But if you do this [sqrt(x)]^2, then my original argument holds, since x is gotta be bigger or equal to 0, hence its not all real x.

Anyways, I was just wondering about this because of the last part of q5 in 1996 hsc, a very good question, have a look if you haven't done it already.
 

Archman

Member
Joined
Jul 29, 2003
Messages
337
Gender
Undisclosed
HSC
N/A
on paper sqrt usually assumes the positive value.
 
C

coca cola

Guest
mojako said:
Induction can be used to prove negative integers.
It should be able to be used to prove non-integers as well, but not all real number.
You can only prove numbers that go with a fixed increment.
For example, prove for n=k and k=k+0.5.
If you can also establish the truth for n=pi,
its true for n=pi, n=pi+0.5, n=pi+1, n=pi+1.5, etc
Technically integers encompasses what you mean, the reason I and books say its non-negative is because we don't count logically with negative objects/integers/entities.
 

Archman

Member
Joined
Jul 29, 2003
Messages
337
Gender
Undisclosed
HSC
N/A
technically it is possible to do inductions on rationals. :p
 

Estel

Tutor
Joined
Nov 12, 2003
Messages
1,261
Gender
Male
HSC
2005
Assume true for n=k
Prove true for lim e--->0 n=k+e
:p
 

BillyMak

Silent majority
Joined
Aug 20, 2004
Messages
443
Location
Randwick
Gender
Male
HSC
2004
x<sup>4/3</sup> = (x<sup>2/3</sup>)<sup>2</sup>

Any real number squared is >=0, since x<sup>2/3</sup> is a real number for any real value of x, x<sup>4/3</sup> >= 0 for all real x.
 

mojako

Active Member
Joined
Mar 27, 2004
Messages
1,333
Gender
Male
HSC
2004
Estel said:
Assume true for n=k
Prove true for lim e--->0 n=k+e
:p
that's if you can actually show that the statement is true for n=k+e :p
this stuff is likely to get super messy (IFF it can be done)
 

Rorix

Active Member
Joined
Jun 29, 2003
Messages
1,818
Gender
Male
HSC
2005
BillyMak said:
x<sup>4/3</sup> = (x<sup>2/3</sup>)<sup>2</sup>

Any real number squared is >=0, since x<sup>2/3</sup> is a real number for any real value of x, x<sup>4/3</sup> >= 0 for all real x.
-2 = (-2)^1 = ((-2)^2)^1/2 = 4^1/2 = 2

cool.
 

dawso

needmorecustomstatusspace
Joined
Feb 25, 2004
Messages
1,029
Gender
Male
HSC
2005
f(x) = x^4/3
f'(x) = 4/3 x ^ 1/3 = 0 when x=0
therefore stat point at (0, 0)
nature - concave up - min
therefore f(x)>=0
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top