[Ewan]

If you have a mapping F: R -> R where

f(x) = [(3x+1)/3] where [,] denotes the "integral part of" - eg

f(x) = [(3x+1)/3] = 3x/3 = x

Is this mapping injective? Now I look at it, and think it is, but my answer [its a previous homework, i'm revising] suggests that it is not injective. Could anyone explain why this is not injective?

[GHOSH-5]

isn't it not injective eg. f(2) = f(7/3)

as injective means

[ghostwalker]

Originally Posted by Ewan

If you have a mapping F: R -> R where

f(x) = [(3x+1)/3] where [,] denotes the "integral part of" - eg

f(x) = [(3x+1)/3] = 3x/3 = x

Is this mapping injective? Now I look at it, and think it is, but my answer [its a previous homework, i'm revising] suggests that it is not injective. Could anyone explain why this is not injective?

Huh!

f(1)=[(3+1)/3]=[4/3]=1
f(1.5)=[(4.5+1)/3]=[5.5/3]=1

[Ewan]

f(1.5) = 1.5?

[ghostwalker]

Originally Posted by Ewan

f(1.5) = 1.5?

Nop. The last thing you do is take the integral part. Your rewriting of the function is incorrect.

[GHOSH-5]

Originally Posted by Ewan

f(1.5) = 1.5?

the floor function surely means f(a) cannot have a fractional/decimal part to it for all a, so f(1.5) certainly isn't 1.5, and as ghostwalker stated, it's 1.

[Ewan]

I'm misunderstanding what "the integral part of" means then. Please explain why f(1.5) =/ 1.5.

I dont understand this part at all f(1.5)=[(4.5+1)/3]=[5.5/3]=1

[GHOSH-5]

Originally Posted by Ewan

I'm misunderstanding what "the integral part of" means then. Please explain why f(1.5) =/ 1.5.

I dont understand this part at all f(1.5)=[(4.5+1)/3]=[5.5/3]=1

the integral part of it means the integer part of a number:

eg [2] = 2, [2.3] = 2, [1.9] = 1, [4.7] = 4, [1/3] = 0, [-2.6] = -3

[Ewan]

Originally Posted by GHOSH-5

the integral part of it means the integer part of a number:

eg [2] = 2, [2.3] = 2, [1.9] = 1, [4.7] = 4, [1/3] = 0

Got cha, cheers.

[Ewan]

Though I'd just add to this thread instead of making a new one...

Either prove or disprove the following statement:
f: A->B
g:B->A

gof = I_A => fog = I_B

[hopinmad]

i take it by I_A you mean the identity function mapping A to A.

so, assume g o f = I: A -> A. then g o f = g(f(a)) = g(b) = a.

so, f o g = f(g(b)) = f(a) = b, so f o g: B -> B is given by (f o g)(b) = b.

so f o g = I_B, as required.