Rational & Irrational Numbers

[nycmath]

Precalculus
David Cohen
Edition 3
Chapter 1, Section 1.1

In exercises 62-64, give an example of irrational numbers a and b such that the indicated expression is (a) rational; (b) irrational.

62. a + b

63. ab

64. a/b

[Eigenvalue]

64. Two square roots whose quotient is a perfect square
[tex]\frac{sqrt{a}}{sqrt{b}}[/tex]=[tex]\sqrt{n²}[/tex]=n

[nycmath]

64. Two square roots whose quotient is a perfect square
[tex]\frac{sqrt{a}}{sqrt{b}}[/tex]=[tex]\sqrt{n²}[/tex]=n

Can you explain your answer to 64?

[Eigenvalue]

a and b are the irrational numbers mentioned in the question.
Let the quotient of a and b be n².
If the quotient of a and b is a perfect square, then the square root of n² will be n, a rational number

If not, the result is irrational

[Eigenvalue]

62. If a and b are opposites, e.g. a=-b or vice versa, the. a-b is rational.
62b. If a is not the opposite of b, then the result is irrational

[nycmath]

a and b are the irrational numbers mentioned in the question.
Let the quotient of a and b be n².
If the quotient of a and b is a perfect square, then the square root of n² will be n, a rational number

If not, the result is irrational

Ok. Now I get it. Thank you.

[nycmath]

62. If a and b are opposites, e.g. a=-b or vice versa, the. a-b is rational.
62b. If a is not the opposite of b, then the result is irrational

Can you give an example using numbers for 62 (a) and (b)?

[Eigenvalue]

62. Let a=[tex]\sqrt{2}[/tex] and b=-[tex]\sqrt{2}[/tex]
a+b=0
0 is rational

If a does not equal b:
Let a=[tex]\sqrt{2}[/tex] and b=[tex]\sqrt{3}[/tex]
a+b=[tex]\sqrt{2}[/tex]+[tex]\sqrt{3}[/tex], which is irrational

[nycmath]

62. Let a=[tex]\sqrt{2}[/tex] and b=-[tex]\sqrt{2}[/tex]
a+b=0
0 is rational

If a does not equal b:
Let a=[tex]\sqrt{2}[/tex] and b=[tex]\sqrt{3}[/tex]
a+b=[tex]\sqrt{2}[/tex]+[tex]\sqrt{3}[/tex], which is irrational

Perfect. You see the importance of using actual numbers in your explanation.

[Eigenvalue]

63. For a to.be rational, a and b must have a product that is a perfect square. If not, it is irrational.

Let a=[tex]\sqrt{2}[/tex] and b=[tex]\sqrt{2}[/tex]
ab=2; 2 is rational

Let a=[tex]\sqrt{2}[/tex] and b=[tex]\sqrt{3}[/tex]
ab=[tex]\sqrt{6}[/tex]; sqrt(6) is irrational

[nycmath]

63. For a to.be rational, a and b must have a product that is a perfect square. If not, it is irrational.

Let a=[tex]\sqrt{2}[/tex] and b=[tex]\sqrt{2}[/tex]
ab=2; 2 is rational

Let a=[tex]\sqrt{2}[/tex] and b=[tex]\sqrt{3}[/tex]
ab=[tex]\sqrt{6}[/tex]; sqrt(6) is irrational

What can I say? To the point and clear.