Sum of two irrational numbers is rational example

We know that quadratic equations can have rational solutions or irrational solutions. For example, the solutions to \((x+3)(x-1)=0\) are -3 and 1, which are rational. The solutions to \(x^2-8=0\) are \(\pm \sqrt{8}\), which are irrational.

Sometimes solutions to equations combine two numbers by addition or multiplication—for example, \(\pm 4\sqrt{3}\) and \(1 +\sqrt {12}\). What kind of number are these expressions?

When we add or multiply two rational numbers, is the result rational or irrational?

• The sum of two rational numbers is rational. Here is one way to explain why it is true:

  • Any two rational numbers can be written \(\frac{a}{b}\) and \(\frac{c}{d}\), where \(a, b, c, \text{ and } d\) are integers, and \(b\) and \(d\) are not zero.
  • The sum of \(\frac{a}{b}\) and \(\frac{c}{d}\) is \(\frac{ad+bc}{bd}\). The denominator is not zero because neither \(b\) nor \(d\) is zero.
  • Multiplying or adding two integers always gives an integer, so we know that \(ad, bc, bd\) and \(ad+bc\) are all integers.
  • If the numerator and denominator of \(\frac{ad+bc}{bd}\) are integers, then the number is a fraction, which is rational.

• The product of two rational numbers is rational. We can show why in a similar way:

  • For any two rational numbers \(\frac{a}{b}\) and \(\frac{c}{d}\), where \(a, b, c, \text{ and } d\) are integers, and \(b\) and \(d\) are not zero, the product is \(\frac{ac}{bd}\).
  • Multiplying two integers always results in an integer, so both \(ac\) and \(bd\) are integers, so \(\frac{ac}{bd}\) is a rational number.

What about two irrational numbers?

• The sum of two irrational numbers could be either rational or irrational. We can show this through examples:

  • \(\sqrt3\) and \(\text-\sqrt3\) are each irrational, but their sum is 0, which is rational.
  • \(\sqrt3\) and \(\sqrt5\) are each irrational, and their sum is irrational.

• The product of two irrational numbers could be either rational or irrational. We can show this through examples:

  • \(\sqrt2\) and \(\sqrt8\) are each irrational, but their product is \(\sqrt{16}\) or 4, which is rational.
  • \(\sqrt2\) and \(\sqrt7\) are each irrational, and their product is \(\sqrt{14}\), which is not a perfect square and is therefore irrational.

What about a rational number and an irrational number?

• The sum of a rational number and an irrational number is irrational. To explain why requires a slightly different argument:

  • Let \(R\) be a rational number and \(I\) an irrational number. We want to show that \(R+I\) is irrational.
  • Suppose \(s\) represents the sum of \(R\) and \(I\) (\(s=R+I\)) and suppose \(s\) is rational.
  • If \(s\) is rational, then \(s + \text-R\) would also be rational, because the sum of two rational numbers is rational.
  • \(s + \text-R\) is not rational, however, because \((R + I) + \text-R = I\).
  • \(s + \text-R\) cannot be both rational and irrational, which means that our original assumption that \(s\) was rational was incorrect. \(s\), which is the sum of a rational number and an irrational number, must be irrational.

• The product of a non-zero rational number and an irrational number is irrational. We can show why this is true in a similar way:

  • Let \(R\) be rational and \(I\) irrational. We want to show that \(R \boldcdot I\) is irrational.
  • Suppose \(p\) is the product of \(R\) and \(I\) (\(p=R \boldcdot I\)) and suppose \(p\) is rational.
  • If \(p\) is rational, then \(p \boldcdot \frac{1}{R}\) would also be rational because the product of two rational numbers is rational.
  • \(p \boldcdot \frac{1}{R}\) is not rational, however, because \(R \boldcdot I \boldcdot \frac{1}{R} = I\).
  • \(p \boldcdot \frac{1}{R}\) cannot be both rational and irrational, which means our original assumption that \(p\) was rational was false. \(p\), which is the product of a rational number and an irrational number, must be irrational.

Can sum of 2 irrational numbers be rational?

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