 Chesley

2020-11-09

Use proof by Contradiction to prove that the sum of an irrational number and a rational number is irrational. Usamah Prosser

A real number r is rational if and only if there exists two integer a and b such that $r=a/b$ ( b is not equal to zero) Let us assume that x is an irrational number and y is a rational number . Suppose for the sake of contradiction that $x+y$ is not irrational $y-a/b$ since s is irrational , x cannot be written as the ratio of two integers Since $x+y$ in not irrational, $x+y$ is rational . By the definition of rational, there exists integers c and d not equal to zero such that $x+y=c/d$

$x+a/b=c/d$
$x=c/d-a/b$
$x=\left(bc-ad\right)/bd$ Since a,b,c,d are integers bc - ad and bd are also integers. Moreover bd is zero as b and d are both non zero. However , this then implies that x is a rational number (as it is written as the ratio of tow integers). which is in contradiction with the fact that with the fact that x is an irrational number. Thus our supposition that $x+y$ is not irrational is false. Which means that $x+y$ is irrational . nick1337

To prove that the sum of an irrational number and a rational number is irrational using proof by contradiction, we assume the opposite, i.e., we assume that the sum of an irrational number $a$ and a rational number $b$ is rational. Let's denote the sum as $c$:
$c=a+b$
Now, since $a$ is irrational and $b$ is rational, we can express $a$ as the difference between $c$ and $b$:
$a=c-b$
We know that the difference between a rational number and a rational number is always rational. Since $b$ is rational and $c$ is the sum of an irrational number and a rational number (which we assumed to be rational), then $c-b$ must also be rational.
However, we have a contradiction since we initially assumed that $a$ is irrational. Thus, our assumption that the sum of an irrational number and a rational number is rational must be false. Therefore, the sum of an irrational number and a rational number is indeed irrational. Vasquez

Step 1:
Let's denote the irrational number as $a$ and the rational number as $b$. We assume that $a+b$ is rational, so we can express it as $a+b=c$, where $c$ is a rational number.
Since $b$ is rational, we can express it as $b=\frac{p}{q}$, where $p$ and $q$ are integers with $q\ne 0$.
Step 2:
Now, let's express $a$ as the sum of its rational and irrational parts: $a=x+y$, where $x$ is the rational part and $y$ is the irrational part.
Substituting these values into the equation $a+b=c$, we have:
$x+y+\frac{p}{q}=c$
Rearranging the terms, we get:
$y=c-x-\frac{p}{q}$
Since $c$, $x$, and $\frac{p}{q}$ are rational numbers (as the sum of rational numbers is rational), the difference $y$ must also be rational.
However, this contradicts the assumption that $y$ is the irrational part of $a$. Therefore, our initial assumption that $a+b$ is rational must be false.
Hence, we conclude that the sum of an irrational number and a rational number is irrational. RizerMix

Let $a$ be an irrational number and $b$ be a rational number. We assume that their sum, $a+b$, is rational.
By definition, a rational number can be expressed as the ratio of two integers. So, we can write $b$ as $b=\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$.
Since $a$ is irrational and $b$ is rational, their sum can be written as $a+b=x$, where $x$ is rational. Now, we can express $a$ as $a=x-b$.
Substituting the values of $a$ and $b$, we have $a=x-\frac{p}{q}$.
Now, we want to show that $a$ is rational, which contradicts the assumption that $a$ is irrational.
Let's consider the expression for $a$ and simplify it:
$a=x-\frac{p}{q}=\frac{qx-p}{q}$
Since $x$ and $p$ are rational numbers, the numerator $qx-p$ is also rational.
Dividing a rational number by a non-zero rational number yields another rational number. Since $q$ is a non-zero integer, the denominator $q$ is a non-zero rational number.
Therefore, $\frac{qx-p}{q}$ is the quotient of two rational numbers and is itself a rational number.
However, this contradicts our assumption that $a$ is irrational. Therefore, our initial assumption that the sum of an irrational number and a rational number is rational must be false.
Hence, we can conclude that the sum of an irrational number and a rational number is irrational.

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