Let’s do convincing arguments about why the sums and also products that rational and also irrational numbers constantly produce specific kinds of numbers.

You are watching: What is the sum of a rational and irrational number


Here space some examples of integers (positive or negative whole numbers):

Experrewildtv.coment with adding any 2 numbers indigenous the perform (or various other integers of her choice). Try to find one or an ext examples of two integers that:

add increase to another integeradd up to a number that is not one integer

Experrewildtv.coment with multiplying any two numbers from the list (or other integers of her choice). Shot to find one or more examples of 2 integers that:

multiply come make another integermultiply to do a number the is not an integer

Here space a couple of examples of adding two reasonable numbers. Is each amount a rational number? Be all set to define 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 sum of 2 rational number is rational.

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

Find the sum of \(\fracab\) and \(\fraccd\). Display your reasoning. In the sum, room the molecule and the denominator integers? how do you know?Use your responses to define why the sum of \(\fracab + \fraccd\) is a rational number. Use the same thinking as in the previous question to describe why the product of two rational numbers, \(\fracab \boldcdot \fraccd\), should be rational.
Consider number that are of the kind \(a + b \sqrt5\), wherein \(a\) and \(b\) are whole numbers. Let’s contact such numbers quintegers.

Here room some instances of quintegers:


When we include two quintegers, will we constantly get one more quinteger? either prove this, or uncover two quintegers whose amount is no a quinteger.When us multiply 2 quintegers, will we constantly get an additional quinteger? one of two people prove this, or discover two quintegers who product is not a quinteger.

Here is a way to describe 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? explain how you know.Evaluate \(s + \text-\frac19\). Is the amount rational or irrational?Use her responses so far to define why \(s\) can not be a rational number, and therefore \( \sqrt2 + \frac 19\) can not be rational.Use the same reasoning as in the earlier question to describe why \(\sqrt2 \boldcdot \frac 19\) is irrational.

Consider the equation \(4x^2 + bx + 9=0\). Find a value of \(b\) so the the equation has:

2 reasonable solutions2 irrational solutions1 solutionno solutionsDescribe every the values of \(b\) that create 2, 1, and no solutions.

Write a new quadratic equation with each form of solution. Be prepared to explain how you understand that her equation has the specified kind and number of solutions.

no solutions2 irrational solutions2 rational solutions1 solution

We know that quadratic equations deserve to have rational remedies or irrational solutions. For example, the options to \((x+3)(x-1)=0\) are -3 and 1, which space rational. The solutions to \(x^2-8=0\) room \(\pm \sqrt8\), which room irrational.

Sometrewildtv.comes services to equations integrate two number by enhancement or multiplication—for example, \(\pm 4\sqrt3\) and \(1 +\sqrt 12\). What sort of number room these expressions?

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

The amount of two rational number is rational. Here is one means to define why it is true:

Any 2 rational numbers can be composed \(\fracab\) and \(\fraccd\), where \(a, b, c, \text and also d\) room integers, and also \(b\) and \(d\) space not zero.The sum of \(\fracab\) and also \(\fraccd\) is \(\fracad+bcbd\). The denominator is not zero due to the fact that neither \(b\) no one \(d\) is zero.Multiplying or adding two integers always gives an integer, therefore we understand that \(ad, bc, bd\) and \(ad+bc\) space all integers.If the numerator and also denominator the \(\fracad+bcbd\) space integers, then the number is a fraction, i beg your pardon is rational.

The product of two rational numbers is rational. Us can present why in a srewildtv.comilar way:

For any type of two rational number \(\fracab\) and \(\fraccd\), wherein \(a, b, c, \text and d\) are integers, and also \(b\) and also \(d\) space not zero, the product is \(\fracacbd\).Multiplying 2 integers always results in one integer, therefore both \(ac\) and \(bd\) are integers, for this reason \(\fracacbd\) is a rational number.

What around two irrational numbers?

The sum of 2 irrational numbers might be one of two people rational or irrational. We can present this through examples:

\(\sqrt3\) and also \(\text-\sqrt3\) space each irrational, however their amount is 0, i beg your pardon is rational.\(\sqrt3\) and also \(\sqrt5\) room each irrational, and their amount is irrational.

The product of 2 irrational numbers can be either rational or irrational. Us can present this with examples:

\(\sqrt2\) and also \(\sqrt8\) room each irrational, yet their product is \(\sqrt16\) or 4, i beg your pardon is rational.\(\sqrt2\) and also \(\sqrt7\) are each irrational, and their product is \(\sqrt14\), which is not a perfect square and also is therefore irrational.

What about a reasonable number and an irrational number?

The amount of a reasonable number and also an irrational number is irrational. To describe why requires a slightly different argument:

Let \(R\) it is in a reasonable number and \(I\) one irrational number. We desire to display that \(R+I\) is irrational.Suppose \(s\) to represent the sum of \(R\) and \(I\) (\(s=R+I\)) and suppose \(s\) is rational.If \(s\) is rational, then \(s + \text-R\) would likewise be rational, because the amount of 2 rational numbers is rational.\(s + \text-R\) is no rational, however, because \((R + I) + \text-R = I\).\(s + \text-R\) cannot be both rational and also irrational, which means that ours original presumption that \(s\) to be rational to be incorrect. \(s\), i m sorry is the amount of a reasonable number and also an irrational number, must be irrational.

The product that a non-zero reasonable number and an irrational number is irrational. Us can present why this is true in a srewildtv.comilar way:

Let \(R\) be rational and also \(I\) irrational. We desire to present that \(R \boldcdot I\) is irrational.Suppose \(p\) is the product of \(R\) and \(I\) (\(p=R \boldcdot I\)) and also suppose \(p\) is rational.If \(p\) is rational, climate \(p \boldcdot \frac1R\) would additionally be rational since the product of 2 rational number is rational.\(p \boldcdot \frac1R\) is no rational, however, due to the fact that \(R \boldcdot ns \boldcdot \frac1R = I\).\(p \boldcdot \frac1R\) cannot be both rational and also irrational, which way our original presumption that \(p\) to be rational to be false. \(p\), which is the product that a reasonable number and an irrational number, have to be irrational.

The formula \(x = \text-b \pm \sqrtb^2-4ac \over 2a\) that offers the solutions of the quadratic equation \(ax^2 + bx + c = 0\), wherein \(a\) is no 0.

See more: Brutus Soliloquy Act 2 Scene 1, No Fear Shakespeare: Julius Caesar: Act 2 Scene 1



The Illustrative mathematics name and logo space not subject to the an innovative Commons license and also may no be supplied without the prior and also express composed consent that Illustrative Mathematics.