## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.12

12. Show that [x] + [x + 1/2] = [2x] whenever x is a real number.

Solution:  We have to prove the following equation
[x]+[x+1/2]=[2x]….(1)

Where x is a real number

Let us consider the LHS of the above equation.

We define

x=[x]+{x}….(2)

Where [x] is the integral part and {x} is the fractional part of x

Now, substituting (2) in (1) will give us
LHS=[[x]+{x}]+[[x]+{x}+1/2]
=[x]+[{x}]+[x]+[{x}+1/2]
We have used the fact that [[x]]=[x] as [x] is itself an integer.

Proceeding forward gives us
LHS=2[x]+[{x}]+[{x}+1/2]
=2[x]+0+[{x}+1/2]
This is because the greatest integer value of any fractional value is zero.

Consider two following cases.

Case 1

We assume that [{x}+1/2]=0

So,
LHS=2[x]+0
=2[x]
=[2x]
This is because 2 is an integer value and can be including in the greatest integer part.

Case 2
We assume that [{x}+1/2]=1

So,
LHS=2[x]+1
=2[x]+[2{x}]
=[[2x]]+[2{x}]
=[2[x]+2{x}]
Therefore,
LHS=[2([x]+{x})]
=[2x]
This we get from (2)

Hence, we have proved the equation (1)

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.11

11. What is the value of[x] + [-x] where x is a real number?
Solution:

We consider the following expression
[x]+[-x]
We consider following two cases

Case 1

If x is an integer, then so is -x

Also,
[x]=x
[-x]=-x
Therefore,
[x]+[-x]=x+(-x)
=x-x
=0

Case 2

If x is not an integer, then so is not -x

Let us take x to be a rational number, that is,
x=1/p
Where $p\neq0,-1,1$

And is an integer

Therefore,
[x]+[-x]=[1/p]+[-1/p]
=0+(-1)
=-1

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.10

10. Find the fractional part of each of these numbers:
(a) -8/5
(b) 22/7
(c) -1
(d) -1/3
Solution:
The fractional part of a real number x, denoted by {x}, is the difference between x and the largest integer less than or equal to x, that is,[x]
In other words, {x}=x-[x]

(a)-8/5

The fractional part of -8/5 is
{-8/5}=-8/5-[-8/5]
=-8/5-(-2)
=(-8+10)/5
=2/5
Therefore,
{-8/5}=2/5

(b) 22/7

The fractional part of 22/7 is
{22/7}=22/7-[22/7]
=22/7-3
=(22-21)/7
=1/7
Therefore,
{22/7}=1/7

(c)-1

The fractional part of -1 is
{-1}=-1-[-1]
=-1-(-1)
=-1+1
=0
Therefore,
{-1}=0

(d)-1/3

The fractional part of -1/3 is
{-1/3}=-1/3-[-1/3]
=-1/3-(-1)
=(-1+3)/3
=2/3
Therefore,
{-1/3}=2/3

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.9

9. Find the fractional part of each of these numbers:
(a) 8/5
(b) 1/7
(c) -11/4
(d) 7

Solution:  The fractional part of a real number x, denoted by {x}, is the difference between x and the largest integer less than or equal to x, that is,[x]

In other words, {x}=x-[x]

(a) 8/5

The fractional part of 8/5 is
{8/5}=8/5-{8/5]
=8/5-1
=(8/5)/5
=3/5
Therefore,
{8/5}=3/5

(b) 1/7

The fractional part of 1/7 is
{1/7}=1/7-[1/7]
=1/7-0
=1/7
Therefore,
{1/7}=1/7

(c) -11/4

The fractional part of -11/4 is
{-11/4}=-11/4-[-11/4]
=-11/4-(-3)
=-11/4+3
=(-11+12)/4
Therefore,
{-11/4}=1/4

(d) 7

The fractional part of 7 is
{7}=7-
=7-7
=0
Therefore,
{7}=0

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.8

8. Find the following values of the greatest integer function.
(a) [-1/4]
(b) [-22/7]
(c) [5/4]
(d) [[1/2]]
(e) [[3/2] + [-3/2]]
(f) [3 – [1/2]]

Solution:  The greatest integer, denoted by [x], is the largest integer less than or equal to x satisfying the inequation
[x] \leq x<[x]+1
Where x is a real number

(a)[-1/4]

The greatest integer less than equal to -1/4 is -1

So, [-1/4]=1

(b) [-22/7]

The greatest integer less than equal to -22/7 is -4

So, [-22/7]=-4

(c) [5/4]

The greatest integer less than equal to 5/4 is 1

So, [5/4]=1

(d) [[1/2]]

The greatest integer less than equal to 1/2 is 0

Now, [[1/2]]=
=0

So, [[1/2]]=0

(e) [[3/2]+[-3/2]]

The greatest integer less than equal to 3/2 is 1

So, [3/2]=1

Also, the greatest integer less than equal to -3/2 is -2

Now,
[[3/2]+[-3/2]]=[1+(-2)]
=[-1]
=-1
So, [[3/2]+[-3/2]]=-1

(f) [3-[1/2]]

The greatest integer less than equal to 1/2 is 0

So, [1/2]=0

Now,
[3-[1/2]]=[3-0]
=
=3
So, [3-[1/2]]=3

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.7

7. Find the following values of the greatest integer function.

a) [1/4]
b) [-3/4]
c) [22/7]
d) [-2]
e) [[1/2]+ [1/2]]
f) [-3+ [-1/2]]

Solution:  The greatest integer, denoted by [x], is the largest integer less than or equal to x satisfying the inequation
$[x]\leq x < [x]+1$
Where x is a real number

(a)[1/4]

The greatest integer less than equal to 1/4 is 0

So, [1/4]=0

(b)[-3/4]

The greatest integer less than equal to -3/4 is -1

So, [-3/4]=-1

(c)[22/7]

The greatest integer less than equal to 22/7 is 3

So, [22/7]=3

(d) [-2]

The greatest integer less than equal to -2 is -2 itself

So,[-2]=-2

(e) [[1/2]+[1/2]]

The greatest integer less than equal to 1/2 is 0

So, [1/2]=0

Now,
[[1/2]+[1/2]]=[0+0]
=
=0
So, [[1/2]+[1/2]]=0

(f) [-3+[-1/2]]

The greatest integer less than equal to -1/2 is -1

So, [-1/2]=-1

Now,
[-3+[-1/2]]=[-3+(-1)]
=[-4]
=-4
So, [-3+[-1/2]]=-4

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.6

6. Show that every nonempty set of negative integers has a greatest element.

Solution: We assume that S is a non-empty set of negative integers.

We now consider the set
$T=\left\{-s:s\in S\right\}$

Now, T is a non-empty set of positive integers and by the well ordering principle has a least element $-s_0$ for some $s_0 \in S$

Then,
$-s_0\leq-s$
For every $s\in S$

Hence,
$s_0\geq s$
For every $s\in S$

This implies that $s_0$ is the greatest element of S.

Therefore, every non-empty set of negative integers has a greatest element.

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.5

5. Use the well-ordering property to show that $\sqrt{3}$ is irrational.

Solution: We prove the statement by contradiction.

Now, we assume that $\sqrt{3}$ is rational.

Then there exist positive integers a and b such that $\sqrt{3}=\frac{a}{b}$

Let us consider the set $S=\left\{k\sqrt{3}:k\ and\ k\sqrt{3}\ are\ positive\ integers\right\}$

The set S is non-empty because $a=b\sqrt{3}$

Therefore, by the well ordering property, has a smallest element, say,

We have,
$s\sqrt{3}-s=s\sqrt{3}-t\sqrt{3}$
$s\sqrt{3}-s=(s-t)\sqrt{3}$

Since, $s\sqrt{3}=3t$ and s are both integers,$(s-t)\sqrt{3}$ must also be an integer.

Also, it is positive, since,
$s\sqrt{3}-s=s(\sqrt{3}-1)$
And
$\sqrt{3}>1$
It is less than s because $s=t\sqrt{3},s\sqrt{3}=3t,\sqrt{3}<3$.
This contradicts our assumption of s being the smallest positive integer in S.
Hence,$\sqrt{3}$ is irrational.

## Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.4

4. Prove or disprove each of the following statements.
a) The sum of a rational and an irrational number is irrational.
b) The sum of two irrational numbers is irrational.
c) The product of a rational number and an irrational number is irrational.
d) The product of two irrational numbers is irrational.

Solution:

(a) The sum of rational and irrational number is irrational. We prove the statement by contradiction. Let us assume that there is a rational number x and an irrational number y such that (x+y) is rational.
Now we define rational numbers as  $x=\frac{a}{b}$…..(1)

For some integers a,b with $b\neq0$ And $x+y=\frac{c}{d}$…..(2)

For some integers c,d with $d\neq0$

we substitute (1) in (2) and get

$\frac{a}{b}+y=\frac{c}{d}$

and so $y=\frac{c}{d}-\frac{a}{b}$ $y=\frac{(bc-ad)}{bd}$

Now (bc-ad) and bd are both integers since products and differences of integers are integers and $bd\neq0$.
Hence, y is a ratio of integers (bc-ad) and bd with $bd\neq0$. So, by the definition of Rational, y is rational. This contradicts the fact that y is irrational. Hence our assumption is wrong and the statement is true.

(b) The sum of two irrational numbers may not necessarily be irrational.
Let us take two irrational numbers $\pi$ and $-\pi$.$\pi$ is irrational as it cannot be expressed as a ratio of two integers.

Summing these irrational numbers gives us $\pi+(-\pi)=\pi-\pi=0$

The number 0 is a rational number. Hence, our statement is justified.

(c) The product of a rational and an irrational number may not necessarily be irrational.

Let us take 0 as a rational number and $\pi$ as an irrational number. We know that any number multiplied by 0 gives 0, that is,
$\pi.0=0$
Hence, our statement is justified.

(d) The product of two irrational numbers may not necessarily be irrational.

Let us take two irrational numbers $\pi$ and $\frac{1}{\pi}$.

Now, $\pi.\frac{1}{\pi}=\frac{\pi}{\pi}=1$

But 1 is rational number. Hence our statement is justified.