In order to prove that the sum of a rational number and an irrational number is irrational, we will use a proof by contraction.
What is a proof by contradiction?
Suppose we want to prove that a math statement is true. The first thing to do is to assume that the math statement is false. Then you need to show that this will lead to a contradiction. If it leads to a contradiction, then the statement must be true.
Assume that the the sum of a rational number and an irrational number is not irrational, but rational.
rational number + irrational number = rational
Let x / y represent the rational number, let a represent the irrational number, and let m / n represent the sum that is "rational"
x / y + a = m / n
Isolate a by subtracting x / y from each side of the equation.
x / y - x / y + a = m / n - x / y
0 + a = m / n - x / y
a = m / n - x / y
Write m / n - x / y with the same denominator.
a = m / n - x / y
The common denominator is ny
Multiply m / n and x / y by ny / ny
a = (m / n)(ny / ny) - (x / y)(ny / ny)
a = (m × ny / (n^{2}y) - (x × ny / (ny^{2})
a = (my / (ny) - (xn / (ny)
a = (my - xn) / (ny)
If x / y and m / n are rational numbers, then we know that x, y, m, and n are integers. This means my, xn, and ny are integers. This will also mean that my - xn is an integer.
Therefore, the expression a = (my - xn) / (ny) is a rational number since it is made by dividing two integers.
This is a contradiction since a is an irrational number as stated above. a cannot be rational and irrational at the same.
Therefore, we can conclude that the sum of a rational number and an irrational number is indeed irrational.
Nov 18, 22 08:20 AM
Nov 17, 22 10:53 AM