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

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Here space some instances of integers (positive or an unfavorable whole numbers):

Expercg-tower.coment with including any two numbers native the list (or other integers of your choice). Try to find one or much more examples of 2 integers that:

add up to one more integeradd as much as a number that is not one integer

Expercg-tower.coment with multiplying any kind of two numbers from the perform (or other integers of your choice). Try to find one or an ext examples of 2 integers that:

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

Here space a few examples of adding two reasonable numbers. Is each amount a reasonable number? Be ready to define how you know.

(4 +0.175 = 4.175)(frac12 + frac45 = frac 510+frac810 = frac1310)( ext-0.75 + frac148 = frac ext-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 two rational number is rational.

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

Find the sum of (fracab) and (fraccd). Present your reasoning. In the sum, room the numerator and the denominator integers? just how do girlfriend know?Use her responses to explain why the amount of (fracab + fraccd) is a rational number. Use the same reasoning as in the previous question to define why the product of two rational numbers, (fracab oldcdot fraccd), need to be rational.
Consider numbers that are of the form (a + b sqrt5), where (a) and (b) are entirety numbers. Let’s speak to such numbers quintegers.

Here are some examples of quintegers:


When we include two quintegers, will we always get an additional quinteger? either prove this, or find two quintegers whose sum is not a quinteger.When we multiply 2 quintegers, will 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 method to explain why (sqrt2 + frac 19) is irrational.

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

Suppose (s) is rational.

Would (s + ext- frac19) be reasonable or irrational? define how friend know.Evaluate (s + ext-frac19). Is the sum rational or irrational?Use your responses so much to describe why (s) can not be a reasonable number, and therefore ( sqrt2 + frac 19) can not be rational.Use the same reasoning as in the earlier question to describe why (sqrt2 oldcdot 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 kind of solution. Be all set to describe how you understand that her equation has actually the specified type and number of solutions.

no solutions2 irrational solutions2 reasonable solutions1 solution

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

Sometcg-tower.comes options to equations integrate two numbers by enhancement or multiplication—for example, (pm 4sqrt3) and (1 +sqrt 12). What type of number space these expressions?

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

The amount of two rational number is rational. Right here is one way to explain why the is true:

Any two rational numbers deserve to be written (fracab) and (fraccd), wherein (a, b, c, ext and also d) space integers, and (b) and also (d) space not zero.The sum of (fracab) and (fraccd) is (fracad+bcbd). The denominator is no zero because neither (b) no one (d) is zero.Multiplying or adding two integers constantly gives one integer, for this reason we understand that (ad, bc, bd) and (ad+bc) are all integers.If the numerator and denominator that (fracad+bcbd) are integers, clcg-tower.comate the number is a fraction, i m sorry is rational.

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

For any type of two rational numbers (fracab) and also (fraccd), where (a, b, c, ext and also d) room integers, and (b) and also (d) are not zero, the product is (fracacbd).Multiplying 2 integers constantly results in an integer, therefore both (ac) and also (bd) space integers, for this reason (fracacbd) is a rational number.

What around two irrational numbers?

The sum of two irrational numbers could be either rational or irrational. Us can show this with examples:

(sqrt3) and ( ext-sqrt3) space each irrational, yet their amount is 0, i beg your pardon is rational.(sqrt3) and (sqrt5) room each irrational, and also their sum is irrational.

The product of two irrational numbers might be either rational or irrational. Us can display this with examples:

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

What about a reasonable number and also an irrational number?

The sum of a reasonable number and also an irrational number is irrational. To define why calls for a slightly various argument:

Let (R) it is in a rational number and also (I) an 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 also suppose (s) is rational.If (s) is rational, clcg-tower.comate (s + ext-R) would also be rational, because the amount of 2 rational number is rational.(s + ext-R) is no rational, however, due to the fact that ((R + I) + ext-R = I).(s + ext-R) cannot be both rational and also irrational, which means that ours original presumption that (s) was rational was incorrect. (s), i m sorry is the amount of a rational number and an irrational number, need to be irrational.

The product the a non-zero reasonable number and an irrational number is irrational. Us can display why this is true in a comparable way:

Let (R) be rational and also (I) irrational. We desire to show that (R oldcdot I) is irrational.Suppose (p) is the product that (R) and also (I) ((p=R oldcdot I)) and also suppose (p) is rational.If (p) is rational, then (p oldcdot frac1R) would also be rational since the product of 2 rational number is rational.(p oldcdot frac1R) is no rational, however, because (R oldcdot i oldcdot frac1R = I).(p oldcdot frac1R) cannot be both rational and irrational, which means our original assumption that (p) to be rational was false. (p), i m sorry is the product of a reasonable number and an irrational number, must be irrational.
Video VLS Alg1U7V5 Rational and also Irrational solutions (Lessons 19–21) accessible at https://player.vcg-tower.comeo.com/video/531442545.

The formula (x = ext-b pm sqrtb^2-4ac over 2a) that gives the options of the quadratic equation (ax^2 + bx + c = 0), where (a) is no 0.

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