You are watching: Is the product of two rational numbers always rational

Here room some instances of integers (positive or negative whole numbers):

Experyellowcomic.coment with including any two numbers from the list (or various other integers of your choice). Shot to discover one or an ext examples of two integers that:

add up to another integeradd as much as a number that is*not*an integer

Experyellowcomic.coment through multiplying any two number from the list (or other integers of her choice). Try to find one or an ext examples of two integers that:

multiply come make one more integermultiply to do a number that is*not*one integer

Here space a couple of examples of including two rational numbers. Is each amount a reasonable number? Be all set to describe how you know.

\(4 +0.175 = 4.175\)\(\frac12 + \frac45 = \frac 510+\frac810 = \frac1310\)\(\text-0.75 + \frac148 = \frac \text-68 + \frac 148 = \frac 88 = 1\)\(a\) is an integer: \(\frac 23+ \frac a15 =\frac1015 + \frac a15 = \frac 10+a15\)Here is a method to describe why the amount of two rational numbers is rational.

Suppose \(\fracab\) and also \(\fraccd\) space fractions. That way that \(a, b, c,\) and \(d\) are integers, and \(b\) and also \(d\) space not 0.

Find the sum of \(\fracab\) and also \(\fraccd\). Present your reasoning. In the sum, are the numerator and the denominator integers? exactly how do you know?Use her responses to define why the sum of \(\fracab + \fraccd\) is a rational number. Use the same thinking as in the previous concern to explain why the product of 2 rational numbers, \(\fracab \boldcdot \fraccd\), need to be rational.Consider number that room of the type \(a + b \sqrt5\), where \(a\) and \(b\) are entirety numbers. Let’s speak to such numbers

*quintegers*.

Here room some examples of quintegers:

When we include two quintegers, will we always get one more quinteger? either prove this, or find two quintegers whose sum is not a quinteger.When us multiply two quintegers, will certainly we constantly get an additional quinteger? one of two people prove this, or uncover two quintegers who product is not a quinteger.

Here is a means to define why \(\sqrt2 + \frac 19\) is irrational.

Let \(s\) it is in the sum of \( \sqrt2\) and also \(\frac 19\), or \(s=\sqrt2 + \frac 19\).

Suppose \(s\) is rational.

Would \(s + \text- \frac19\) be rational or irrational? describe how friend know.Evaluate \(s + \text-\frac19\). Is the amount rational or irrational?Use your responses so much to describe why \(s\) can not be a rational number, and therefore \( \sqrt2 + \frac 19\) cannot be rational.Use the same thinking as in the earlier question to define why \(\sqrt2 \boldcdot \frac 19\) is irrational.Consider the equation \(4x^2 + bx + 9=0\). Find a value of \(b\) so that the equation has:

2 rational solutions2 irrational solutions1 solutionno solutionsDescribe every the worths of \(b\) that develop 2, 1, and also no solutions.Write a new quadratic equation v each type of solution. Be prepared to explain how you understand that your equation has the specified kind and variety of solutions.

no solutions2 irrational solutions2 rational solutions1 solutionWe know that quadratic equations can have rational options or irrational solutions. For example, the services to \((x+3)(x-1)=0\) are -3 and also 1, which are rational. The options to \(x^2-8=0\) are \(\pm \sqrt8\), which room irrational.

Sometyellowcomic.comes solutions to equations integrate two number by addition or multiplication—for example, \(\pm 4\sqrt3\) and also \(1 +\sqrt 12\). What type of number are these expressions?

When we add or multiply 2 rational numbers, is the an outcome rational or irrational?

The sum of two rational numbers is rational. Below is one method to define why it is true:

Any 2 rational numbers deserve to be written \(\fracab\) and also \(\fraccd\), where \(a, b, c, \text and also d\) space integers, and also \(b\) and \(d\) room not zero.The sum of \(\fracab\) and also \(\fraccd\) is \(\fracad+bcbd\). The denominator is no zero due to the fact that neither \(b\) no one \(d\) is zero.Multiplying or adding two integers constantly gives one integer, so we understand that \(ad, bc, bd\) and also \(ad+bc\) space all integers.If the numerator and also denominator of \(\fracad+bcbd\) are integers, then the number is a fraction, i m sorry is rational.The product of two rational numbers is rational. Us can present why in a syellowcomic.comilar way:

For any type of two rational numbers \(\fracab\) and \(\fraccd\), whereby \(a, b, c, \text and also d\) are integers, and also \(b\) and also \(d\) space not zero, the product is \(\fracacbd\).Multiplying 2 integers constantly results in an integer, for this reason both \(ac\) and also \(bd\) space integers, for this reason \(\fracacbd\) is a reasonable number.What about two irrational numbers?

The amount of 2 irrational numbers can be one of two people rational or irrational. Us can show this with examples:

\(\sqrt3\) and also \(\text-\sqrt3\) space each irrational, but their sum is 0, which is rational.\(\sqrt3\) and \(\sqrt5\) space each irrational, and their sum is irrational.The product of 2 irrational numbers might be one of two people rational or irrational. We can display this with examples:

\(\sqrt2\) and also \(\sqrt8\) space each irrational, but their product is \(\sqrt16\) or 4, which is rational.\(\sqrt2\) and \(\sqrt7\) space each irrational, and their product is \(\sqrt14\), which is no a perfect square and is thus irrational.What about a reasonable number and an irrational number?

The amount of a reasonable number and an irrational number is irrational. To explain why needs a slightly different argument:

Let \(R\) be a reasonable number and \(I\) one irrational number. We want to show that \(R+I\) is irrational.Suppose \(s\) represents the amount of \(R\) and also \(I\) (\(s=R+I\)) and also suppose \(s\) is rational.If \(s\) is rational, climate \(s + \text-R\) would also be rational, since the amount of two rational numbers is rational.\(s + \text-R\) is not rational, however, due to the fact that \((R + I) + \text-R = I\).\(s + \text-R\) cannot be both rational and also irrational, which means that our original assumption that \(s\) was rational to be incorrect. \(s\), i m sorry is the amount of a rational number and also an irrational number, need to be irrational.The product of a non-zero reasonable number and an irrational number is irrational. Us can display why this is true in a syellowcomic.comilar way:

Let \(R\) it is in rational and \(I\) irrational. We want to show that \(R \boldcdot I\) is irrational.Suppose \(p\) is the product of \(R\) and also \(I\) (\(p=R \boldcdot I\)) and also suppose \(p\) is rational.If \(p\) is rational, then \(p \boldcdot \frac1R\) would likewise be rational because the product of two rational numbers is rational.\(p \boldcdot \frac1R\) is no rational, however, because \(R \boldcdot i \boldcdot \frac1R = I\).\(p \boldcdot \frac1R\) cannot be both rational and irrational, which means our original assumption that \(p\) to be rational was false. \(p\), i beg your pardon is the product of a reasonable number and an irrational number, have to be irrational.Video

*VLS Alg1U7V5 Rational and Irrational remedies (Lessons 19–21)*obtainable at https://player.vyellowcomic.comeo.com/video/531442545.

The formula \(x = \text-b \pm \sqrtb^2-4ac \over 2a\) that gives the options of the quadratic equation \(ax^2 + bx + c = 0\), where \(a\) is not 0.

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