32. Show that f: R -> R defined as f(x) = x / sqrt(1 + $x^2$) is one-one but not onto. (function, one-one, onto, real numbers, domain, codomain, range)

Step 1: Define the function

We are given the function $f(x)$ mapping real numbers to real numbers. We need to demonstrate that this function is one-one (injective) but not onto (surjective).

Step 2: Check for one-one property

To prove that a function is one-one, we assume that $f(x_1) = f(x_2)$ for two elements $x_1, x_2$ in the domain. If this assumption leads to $x_1 = x_2$, then the function is one-one.

Step 3: Set up the equality for one-one proof

Substitute the function definition into the assumption $f(x_1) = f(x_2)$. We need to determine if this equation implies $x_1 = x_2$. Remember that the sign of $\sqrt{1+x^2}$ is always positive.

Step 4: Square both sides

To eliminate the square roots, we square both sides of the equation. This simplifies the expression and allows us to manipulate it further algebraically.

Step 5: Rearrange and simplify

Cross-multiply and expand the terms. Notice that the term $x_1^2x_2^2$ appears on both sides, cancelling out. This leaves us with $x_1^2 = x_2^2$.

Step 6: Consider square root carefully

From $x_1^2 = x_2^2$, it implies that $x_1 = \pm x_2$. We need to show that $x_1$ and $x_2$ must have the same sign for $f(x_1) = f(x_2)$ to hold.

Step 7: Conclude one-one property

If $x_1 = -x_2$ (and $x_1 \neq 0$), then $f(x_1) = \frac{x_1}{\sqrt{1+x_1^2}}$ and $f(x_2) = \frac{-x_1}{\sqrt{1+(-x_1)^2}} = \frac{-x_1}{\sqrt{1+x_1^2}} = -f(x_1)$. If $f(x_1)=f(x_2)$, then $f(x_1)=-f(x_1)$, which means $2f(x_1)=0$, so $f(x_1)=0$. This implies $x_1=0$. Therefore, if $f(x_1)=f(x_2)$, we must have $x_1=x_2$. Hence, the function is one-one.

Step 8: Check for onto property

To check if the function is onto, we need to find the range of $f(x)$ and determine if it equals the codomain, which is $\mathbb{R}$ (all real numbers). We set $y=f(x)$ and try to express $x$ in terms of $y$.

Step 9: Solve for x in terms of y

Square both sides to eliminate the square root. Then, rearrange the equation to isolate $x^2$. This allows us to determine the possible values of $y$ for which $x$ can be a real number.

Step 10: Determine the range of y

For $x$ to be a real number, $x^2$ must be non-negative. This means $\frac{y^2}{1 - y^2}$ must be greater than or equal to 0. Since $y^2 \geq 0$, we require $1 - y^2 > 0$. This inequality implies that $-1 < y < 1$ for all values in the range.

Step 11: Conclude onto property

The range of the function is the interval $(-1, 1)$, which means that for any real number $y$ outside this interval (e.g., $y=2$), there is no real $x$ such that $f(x)=y$. Since the range is not equal to the codomain $\mathbb{R}$, the function is not onto.

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