2.3 For each of the following equations, find an integer x that satisfies the equation.a. 5x = 4(mod 3)b. 7x = 6(mod 5)c. 9x = 8(mod 7)
2.5 A modulus of 0 does not fit the definition but is defined by convention as follows:a mod 0 = a. With this definition in mind, what does the following expression mean:a = b(mod 0)?
The Correct Answer and Explanation is :
Let’s go through each of the equations in part 2.3 and solve them first, then I will explain the modulus of 0 and its implications in part 2.5.
2.3 Solving for x in modular equations:
a) ( 5x = 4 \mod 3 )
We can reduce the equation modulo 3:
- ( 5 \equiv 2 \mod 3 ) (since 5 divided by 3 gives a remainder of 2)
- So, the equation becomes:
[ 2x \equiv 4 \mod 3 ]
Now, reduce 4 modulo 3:
[ 4 \equiv 1 \mod 3 ]
So, the equation becomes:
[ 2x \equiv 1 \mod 3 ]
Now, we need to find the integer value of ( x ) that satisfies this. To do this, find the multiplicative inverse of 2 modulo 3, which is a number ( y ) such that ( 2y \equiv 1 \mod 3 ). We test:
- ( 2 \times 2 = 4 \equiv 1 \mod 3 )
So, the multiplicative inverse of 2 modulo 3 is 2. Multiply both sides of the equation ( 2x \equiv 1 \mod 3 ) by 2:
[
2 \times 2x \equiv 2 \times 1 \mod 3
]
[
4x \equiv 2 \mod 3
]
Since ( 4 \equiv 1 \mod 3 ), this simplifies to:
[
x \equiv 2 \mod 3
]
Thus, ( x = 2 ) satisfies the equation.
b) ( 7x = 6 \mod 5 )
We reduce both sides modulo 5:
- ( 7 \equiv 2 \mod 5 )
- ( 6 \equiv 1 \mod 5 )
So the equation becomes:
[
2x \equiv 1 \mod 5
]
Next, find the multiplicative inverse of 2 modulo 5. We test:
- ( 2 \times 3 = 6 \equiv 1 \mod 5 )
So, multiply both sides of the equation by 3:
[
3 \times 2x \equiv 3 \times 1 \mod 5
]
[
6x \equiv 3 \mod 5
]
Since ( 6 \equiv 1 \mod 5 ), this simplifies to:
[
x \equiv 3 \mod 5
]
Thus, ( x = 3 ) satisfies the equation.
c) ( 9x = 8 \mod 7 )
We reduce both sides modulo 7:
- ( 9 \equiv 2 \mod 7 )
- ( 8 \equiv 1 \mod 7 )
So the equation becomes:
[
2x \equiv 1 \mod 7
]
Next, find the multiplicative inverse of 2 modulo 7. We test:
- ( 2 \times 4 = 8 \equiv 1 \mod 7 )
So, multiply both sides of the equation by 4:
[
4 \times 2x \equiv 4 \times 1 \mod 7
]
[
8x \equiv 4 \mod 7
]
Since ( 8 \equiv 1 \mod 7 ), this simplifies to:
[
x \equiv 4 \mod 7
]
Thus, ( x = 4 ) satisfies the equation.
2.5 Meaning of ( a = b \mod 0 ):
The expression ( a = b \mod 0 ) is undefined under normal modular arithmetic, because the modulus operation involves division, and division by zero is undefined. However, in some contexts, it’s defined by convention as ( a \mod 0 = a ), meaning that the result of the operation does not change the value of ( a ).
This convention is useful when we deal with edge cases or when the modulus operation is extended beyond its normal range of validity. Essentially, it means that any number is congruent to itself when taken modulo zero. This might seem paradoxical because modulus typically requires a non-zero divisor to perform meaningful division, but this rule serves to provide consistency in some mathematical systems and to handle edge cases.
However, from a theoretical standpoint, such a definition doesn’t follow the usual arithmetic rules and is considered non-standard. It is typically only encountered in particular branches of mathematics or specialized applications where such definitions are chosen for convenience.
Now, I’ll generate an image to illustrate this concept.
Here is an educational illustration that explains the concept of modular arithmetic with a modulus of 0. It visualizes the equation ( a = b \mod 0 ) and provides a simple explanation that modulo 0 is defined conventionally as ( a = a ). This image should help clarify how this non-standard rule works in certain mathematical contexts.