Q.Let $A = \{1, 2, 3\}$, $B = \{4, 5, 6, 7\}$ and let $f = \{(1, 4), (2, 5), (3, 6)\}$ be a function from A to B. Show that $f$ is one-one.

The statement given is the textbook definition of a one-one (injective) function.

A function f : A → B is called one-one (or injective) if distinct elements of A always map to distinct elements of B — equivalently, if f(a) = f(b) forces a = b (no two different inputs can share an output).

This is precisely the condition stated in the question, so f is one-one.

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Exhibit a counterexample: two distinct inputs with the same image.

A function $f$ is one-one (injective) if $f(x_1)=f(x_2)\Rightarrow x_1=x_2$. To show $f(x)=x^2$ is not one-one, we produce two distinct inputs with equal outputs.

Take $x_1=1$ and $x_2=-1$. These are distinct: $1\ne-1$. But

$f(1)=1^2=1,\qquad f(-1)=(-1)^2=1,$

so $f(1)=f(-1)$.

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Check A (is $f$ one-one?) and R (is $f$ onto?) independently, then judge whether R explains A.

Checking Assertion (A): $f:\mathbf{Z}\to\mathbf{Z}$, $f(x)=2x$. If $f(x_1)=f(x_2)$, then $2x_1=2x_2 \Rightarrow x_1=x_2$. So $f$ is one-one. A is TRUE.

Checking Reason (R): The range of $f$ is $\{\ldots,-4,-2,0,2,4,\ldots\}$, i.e. only even integers. Odd integers in the codomain $\mathbf{Z}$ (e.g. 1) have no pre-image. So $f$ is not onto. R is TRUE.

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The stated condition is the wrong implication — it should read $f(x_1)=f(x_2)\Rightarrow x_1=x_2$.

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For injectivity, show $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ using uniqueness of real cube roots.

Let $f(x)=x^3$ and suppose $f(x_1)=f(x_2)$ for $x_1,x_2\in R$. Then

$x_1^3=x_2^3 \Rightarrow x_1^3-x_2^3=0 \Rightarrow (x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0.$

The quadratic factor $x_1^2+x_1x_2+x_2^2=\left(x_1+\tfrac{x_2}{2}\right)^2+\tfrac{3}{4}x_2^2\ge0$, and it is zero only when $x_1=x_2=0$. In every case we are forced to $x_1=x_2$. Hence $f$ is one-one (injective).

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A function is many-one if two different inputs give the same output; $f(x)=x^2$ does this.

A function $f$ is one-one (injective) if distinct elements of the domain always map to distinct elements of the codomain; otherwise it is many-one.

Take $x_1=1$ and $x_2=-1$ in $\mathbf{R}$. Then

$f(1)=1^2=1,\qquad f(-1)=(-1)^2=1$ …

$f^{-1}(17)$ means: find all $x$ with $f(x)=17$.

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To show $f$ is not one-one (not injective), it suffices to exhibit two distinct inputs with the same output.

Given $f:\mathbb R\to\mathbb R$, $f(x)=x^2$ for all $x\in\mathbb R$.

A function is one-one if $f(x_1)=f(x_2) \implies x_1=x_2$ for all $x_1,x_2$ in the domain. We look for a counterexample.

Take $x_1 = 1$ and $x_2 = -1$. These are distinct: $1 \neq -1$.

Compute:

$f(1) = 1^2 = 1, \qquad f(-1) = (-1)^2 = 1$

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The function is one-one (injective).

A function $f$ is one-one (injective) if distinct inputs give distinct outputs, equivalently $f(x_1)=f(x_2)\Rightarrow x_1=x_2$. This is precis …