Name: Algebraic Operations: Functions
Uploaded: 2019-07-19
Duration: 7 min 10 s
Description: Now that we know what functions are, what can we do with them? Can we do operations with them? Yes! What won't these functions do? Let us learn more here.

Question

Find the integral solutions to the equation

[x] [y] = x + y .

Show that all the non-integral solutions lie on exactly two lines. Determine these lines. Here

[.]

denotes the greatest integer function.

Accepted Answer

Given, $[x] [y] = x + y$ , for $x, y \in I$

So, $x y = x + y$

By hit and trial method, we get integral solution as

$x = y = 0$ and $x = y = 2$

Now, when $x, y$ are non-integers

Let $x = I_{1} + f_{1}$ when $I_{1}$ is an integer and $f_{1}$ is the fractional part of $x$ and, $y = I_{2} + f_{2}$ , when $I_{2}$ is an integer and $f_{2}$ is the fractional part of $y$

$\Rightarrow [x] . [y] = x + y$

$\Rightarrow [I_{1} + f_{1}] . [I_{2} + f_{2}] = (I_{1} + f_{1}) + (I_{2} + f_{2})$

$\Rightarrow I_{1} I_{2} = (I_{1} + I_{2}) + (f_{1} + f_{2}) . . . (i)$

$\Rightarrow f_{1} + f_{2} = I_{1} . I_{2} - (I_{1} + I_{2})$

Since, $I_{1} \cdot I_{2}$ is an integer and $(I_{1} + I_{2})$ is integer, therefore, $((I_{1} \cdot I_{2}) - (I_{1} + I_{2}))$ will be an integer.

So, $f_{1} + f_{2}$ is an integer

We know that fractional part of any number lies between zero and one

$\Rightarrow 0 < f_{1} < 1$ and $0 < f_{2} < 1$

$\Rightarrow 0 < f_{1} + f_{2} < 2$

So, the only possible integral value of $f_{1} + f_{2}$ is $1$ .

$\Rightarrow f_{1} + f_{2} = 1$

Putting $f_{1} + f_{2} = 1$ in equation $(i)$ , we get

So, $I_{1} . I_{2} = I_{1} + I_{2} + 1 . . . (A)$

$\Rightarrow I_{1} . I_{2} - I_{1} - I_{2} = 1$

Adding $1$ on both the sides

$\Rightarrow I_{1} . I_{2} - I_{1} - I_{2} + 1 = 2$

$\Rightarrow I_{1} (I_{2} - 1) - 1 (I_{2} - 1) = 2$

$\Rightarrow (I_{1} - 1) (I_{2} - 1) = 2$

Here, $I_{1} - 1$ and $I_{2} - 1$ is an integer

So, here product of two integers is $2$ , so the possible value of $(I_{1} - 1)$ and $(I_{2} - 1)$ will be,

$(I_{1} - 1) = 1, (I_{2} - 1) = 2 . . . (ii)$

or $(I_{1} - 1) = 2, (I_{2} - 1) = 1 . . . (iii)$

We can also take,

$(I_{1} - 1) = - 1, (I_{2} - 1) = - 2 . . . (iv)$

or $(I_{1} - 1) = - 2, (I_{2} - 1) = 1 . . . (v)$

From equations $(ii)$ and $(iii)$ , we get,

$I_{1} = 2$ and $I_{2} = 3$

Or $I_{1} = 3$ and $I_{2} = 2$

As given $[x] [y] = x + y$

$I_{1} . I_{2} = x + y$

$\Rightarrow x + y = 6$

From equations $(iv)$ and $(v)$ , we get

$I_{1} = 0 & I_{2} = - 1$

Or $I_{1} = - 1$ & $I_{2} = 0$

As given, $[x] . [y] = x + y$

$\Rightarrow I_{1} . I_{2} = x + y$

$\Rightarrow - 1.0 = x + y$

$\Rightarrow x + y = 0$

Hence, the non-integral solution lies on the line

$x + y = 6$ and $x + y = 0$

Find the integral solutions to the equation $[x] [y] = x + y .$ Show that all the non-integral solutions lie on exactly two lines. Determine these lines. Here $[.]$ denotes the greatest integer function.

Important Questions on Functions

Learn and Prepare for any exam you want !

Find the integral solutions to the equation [x][y]=x+y. Show that all the non-integral solutions lie on exactly two lines. Determine these lines. Here [.] denotes the greatest integer function.

Important Questions on Functions

Learn and Prepare for any exam you want !

Find the integral solutions to the equation $[x] [y] = x + y .$ Show that all the non-integral solutions lie on exactly two lines. Determine these lines. Here $[.]$ denotes the greatest integer function.