Help: with a Function problem

Hello, i have a few questions regarding functions, its kinda urgent hope you can help me undersand;

Consider N0 as the set of naturals including 0, that is, N0 = N ∪ {0}, and consider the function: N0 → N0, ​​defined by F (n) = n + (−1) ^ n

  1. Is F surjective?

  2. Is F injective?

  3. Consider the following S relation in N0 × N0. ∀a, b ∈ N0, [a S b] ⇔ F (a) = F (b) Show that S is an equivalence relation on N0, and determine the equivalence class of a = 0.

Hope you guys can help me.

Functions #HelpMe! #Advice #Math #Urgent #Please

#Algebra

Note by Tati Barrera
1 month ago

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Comments

Let y=F(x)y=F(x) ⇒y=x+(−1)x\Rightarrow y= x + (-1)^x y≡1( mod 2)⇒x≡0( mod 2)⇒x+1=y⇒x=y−1y\equiv 1 (\bmod 2)\Rightarrow x\equiv 0 (\bmod 2) \Rightarrow x+1=y\Rightarrow \boxed{x=y-1} y≡0( mod 2)⇒x≡1( mod 2)⇒x−1=y⇒x=y+1y\equiv 0 (\bmod 2)\Rightarrow x\equiv 1 (\bmod 2) \Rightarrow x-1=y\Rightarrow \boxed{x=y+1} ∀y∈N∪{0}∃!x∈N∪{0}:F(x)=y\forall y\in \mathbb{N} \cup \{0\} \exists! x\in\mathbb{N}\cup \{0\}:F(x)=y Hence we proved that F(x)F(x) is surjective and injective ∀a,b∈N0×N0,aSb⇒F(a)=F(b)\forall a,b\in N_0 \times N_0, aSb\Rightarrow F(a)=F(b) ∵∀a∈N0,F(a)=F(a)⇒aSa\because \forall a\in N_0, F(a)=F(a)\Rightarrow aSa ∵∀a,b∈N0,F(a)=F(b)⇒F(b)=F(a)∴(aSb⇒bSa)\because \forall a,b\in N_0, F(a)=F(b)\Rightarrow F(b)=F(a)\therefore (aSb\Rightarrow bSa) ∵∀a,b,c∈N0,(F(a)=F(b))∧(F(b)=F(c))⇒F(a)=F(c)⇒(aSb∧bSc⇒aSc)\because \forall a,b,c\in N_0 ,(F(a)=F(b))\wedge (F(b)=F(c))\Rightarrow F(a)=F(c)\Rightarrow (aSb\wedge bSc\Rightarrow aSc) Hence we SS is a equivalence relation [a]={x∈N0:xSa}={x∈N0:F(x)=F(a)}={a}[a]=\{x\in N_0:xSa\}=\{x\in N_0:F(x)=F(a)\}=\{a\} ⇒[0]={0}\Rightarrow [0]=\{0\}

Zakir Husain - 1 month ago

@Zakir Husain

Jeff Giff - 1 month ago
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