Prove that the function $f:N\to N$ defined by $f(x)=x-1$, when $x>2$ and $f(1)=f(2)=1$ is onto but it is not one-one.

> > [!TLDR] > Each natural number is hit (onto), yet $1$ and $2$ share the image $1$ (not one-one). > > **Concept.** *Onto (surjective)*: every element of the codomain has at least one pre-image. *One-one (injective)*: distinct inp …

Prove that the function f:N→ N defined by f(x)=x-1

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f:N\to N

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2023Subjective· 1mImportance★★★★★est

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Every $y\in N$ has a pre-image ( $y=1$ from $f(1)$ ; any $y\ge1$ fro …

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Question of 99

f:N\to N

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2023Subjective· 1mImportance★★★★★est

0% · 0/99 Questions

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Every $y\in N$ has a pre-image ( $y=1$ from $f(1)$ ; any $y\ge1$ fro …

Q.Prove that the function $f:N\to N$ defined by $f(x)=x-1$ , when $x>2$ and $f(1)=f(2)=1$ is onto but it is not one-one.

Q.Prove that the function $f:N\to N$ defined by $f(x)=x-1$ , when $x>2$ and $f(1)=f(2)=1$ is onto but it is not one-one.

Q.Prove that the function f:N→Nf:N\to Nf:N→N defined by f(x)=x−1f(x)=x-1f(x)=x−1, when x>2x>2x>2 and f(1)=f(2)=1f(1)=f(2)=1f(1)=f(2)=1 is onto but it is not one-one.

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Q.Prove that the function f:N→Nf:N\to Nf:N→N defined by f(x)=x−1f(x)=x-1f(x)=x−1, when x>2x>2x>2 and f(1)=f(2)=1f(1)=f(2)=1f(1)=f(2)=1 is onto but it is not one-one.

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Q.Prove that the function $f:N\to N$ defined by $f(x)=x-1$ , when $x>2$ and $f(1)=f(2)=1$ is onto but it is not one-one.

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Q.Prove that the function $f:N\to N$ defined by $f(x)=x-1$ , when $x>2$ and $f(1)=f(2)=1$ is onto but it is not one-one.

Unlock the full solution