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Here room some examples of integers (positive or an unfavorable whole numbers):

Experpiersonforcongress.coment with including any two numbers indigenous the perform (or various other integers of your choice). Try to find one or more examples of 2 integers that:

add up to one more integeradd approxpiersonforcongress.comately a number the is*not*an integer

Experpiersonforcongress.coment with multiplying any kind of two numbers from the list (or various other integers of your choice). Shot to find one or an ext examples of two integers that:

multiply come make an additional integermultiply to make a number the is*not*an integer

Here are a couple of examples of including two rational numbers. Is each sum a reasonable number? Be ready to define how girlfriend 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 2 rational numbers is rational.

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

Find the amount of (fracab) and (fraccd). Display your reasoning. In the sum, room the molecule and the denominator integers? just how do friend know?Use your responses to explain why the sum of (fracab + fraccd) is a rational number. Use the same thinking as in the previous inquiry to explain why the product of two rational numbers, (fracab oldcdot fraccd), have to be rational.Consider numbers that space of the form (a + b sqrt5), where (a) and (b) are whole numbers. Let’s call such numbers

*quintegers*.

Here space some examples of quintegers:

When we include two quintegers, will certainly we constantly get another quinteger? either prove this, or uncover two quintegers whose amount is not a quinteger.When we multiply two quintegers, will we always get another quinteger? either prove this, or discover two quintegers whose product is not a quinteger.

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

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

Suppose (s) is rational.

Would (s + ext- frac19) be reasonable or irrational? describe how friend know.Evaluate (s + ext-frac19). Is the amount rational or irrational?Use your responses so far to explain why (s) cannot be a reasonable number, and also therefore ( sqrt2 + frac 19) cannot 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 worth of (b) so the the equation has:

2 reasonable solutions2 irrational solutions1 solutionno solutionsDescribe all the worths of (b) that develop 2, 1, and also no solutions.Write a new quadratic equation with each form of solution. Be ready to define how you recognize that your equation has actually the specified form and number of solutions.

no solutions2 irrational solutions2 reasonable solutions1 solutionWe understand that quadratic equations have the right to have rational solutions or irrational solutions. Because that example, the options to ((x+3)(x-1)=0) room -3 and 1, which space rational. The services to (x^2-8=0) room (pm sqrt8), which room irrational.

Sometpiersonforcongress.comes remedies to equations incorporate two numbers by addition or multiplication—for example, (pm 4sqrt3) and also (1 +sqrt 12). What type of number space these expressions?

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

The sum of 2 rational number is rational. Below is one means to describe why that is true:

Any 2 rational numbers can be written (fracab) and also (fraccd), wherein (a, b, c, ext and also d) room integers, and (b) and also (d) space not zero.The sum of (fracab) and (fraccd) is (fracad+bcbd). The denominator is no zero since neither (b) nor (d) is zero.Multiplying or including two integers always gives one integer, so we know that (ad, bc, bd) and (ad+bc) are all integers.If the numerator and denominator of (fracad+bcbd) room integers, then the number is a fraction, i beg your pardon is rational.The product of two rational number is rational. We can present why in a spiersonforcongress.comilar way:

For any two rational number (fracab) and (fraccd), wherein (a, b, c, ext and also d) room integers, and also (b) and (d) room not zero, the product is (fracacbd).Multiplying 2 integers always results in an integer, so both (ac) and (bd) room integers, so (fracacbd) is a reasonable number.What around two irrational numbers?

The amount of 2 irrational numbers might be either rational or irrational. We can show this through examples:

(sqrt3) and ( ext-sqrt3) space each irrational, however their sum is 0, which is rational.(sqrt3) and (sqrt5) are each irrational, and their amount is irrational.The product of 2 irrational numbers might be either rational or irrational. Us can display this v examples:

(sqrt2) and also (sqrt8) room each irrational, but their product is (sqrt16) or 4, i m sorry is rational.(sqrt2) and (sqrt7) space each irrational, and also their product is (sqrt14), i beg your pardon is no a perfect square and also is therefore irrational.What around a rational number and also an irrational number?

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

Let (R) be a reasonable number and also (I) an irrational number. We desire to show that (R+I) is irrational.Suppose (s) represents the sum of (R) and (I) ((s=R+I)) and also suppose (s) is rational.If (s) is rational, clpiersonforcongress.comate (s + ext-R) would also be rational, due to the fact that the sum of two rational number is rational.(s + ext-R) is not rational, however, because ((R + I) + ext-R = I).(s + ext-R) cannot be both rational and also irrational, which way that our original presumption that (s) was rational was incorrect. (s), i m sorry is the sum of a reasonable number and also an irrational number, should be irrational.The product that a non-zero rational number and an irrational number is irrational. We can present why this is true in a comparable way:

Let (R) be rational and (I) irrational. We want to display that (R oldcdot I) is irrational.Suppose (p) is the product the (R) and (I) ((p=R oldcdot I)) and suppose (p) is rational.If (p) is rational, then (p oldcdot frac1R) would also be rational because the product of two rational numbers is rational.(p oldcdot frac1R) is no rational, however, because (R oldcdot ns oldcdot frac1R = I).(p oldcdot frac1R) can not be both rational and also irrational, which way our original presumption that (p) to be rational was false. (p), i beg your pardon is the product the a rational number and an irrational number, should be irrational.Video

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

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

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