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]
=7-7
=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]
=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]
=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]
=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.