Respuesta :
1. From your description, I can infer that the multiplication is:
[tex] \frac{ \sqrt{196} }{7} * \sqrt{108} [/tex]
The first thing we are going to do is simplify the radicands 196 ans 108 (picture 1):
[tex]196=2^2*7^2[/tex] and [tex]108=2^2*3^3[/tex]
Knowing this, we can rewrite our radicals as follows:
[tex]\frac{ \sqrt{196} }{7} * \sqrt{108}= \frac{ \sqrt{2^2*7^2} }{7} * \sqrt{2^2*3^3} [/tex]
Remember that [tex] \sqrt[n]{x^n} =x[/tex]; in other words if the radicand is raised to the same power as the index of the radical, we can take the radicand out. Since 2 and 7 are raised to the power 2 and the index of the radical is also 2 (square root), we can take out 2 and 7:
[tex]\frac{ \sqrt{2^2*7^2} }{7} * \sqrt{2^2*3^3}= \frac{2*7}{7} *2 \sqrt{3^3} [/tex]
Look! we have the same numerator and denominator in our fraction, so we can cancel them both:
[tex]\frac{2*7}{7} *2 \sqrt{3^3}=2*2 \sqrt{3^3} =4 \sqrt{3^3} [/tex]
Notice that we can write [tex]3^3[/tex] as [tex]3^2*3[/tex], so we can rewrite our expression one last time:
[tex]4 \sqrt{3^3} =4 \sqrt{3^2*3} =4*3 \sqrt{3} =12 \sqrt{3} [/tex]
We can conclude that the correct option is: [tex]12 \sqrt{3} [/tex]
2. The product of a nonzero rational number and an irrational number is always an irrational number.
Proof by contradiction:
Lets assume that the product of an irrational number and a rational non-zero number is always rational.
Let [tex]x[/tex] be and irrational number and let [tex] \frac{a}{b} [/tex] and [tex] \frac{c}{d} [/tex] be two rational numbers with [tex]a[/tex], [tex]b[/tex], [tex]c[/tex], and [tex]d[/tex] are non-zero integers.
[tex]x* \frac{a}{b} = \frac{c}{d} [/tex]
[tex]x= \frac{c}{d} * \frac{b}{a} [/tex]
[tex]x=\frac{cb}{da}[/tex]
Since integers are closed under multiplication, [tex] \frac{cb}{da}[/tex] is a rational number. Since[tex]x[/tex] is an irrational number and [tex]x=\frac{cb}{da}[/tex], we have a logical contradiction, so we can conclude that the product of an irrational number and a rational non-zero number is always an irrational number.
[tex] \frac{ \sqrt{196} }{7} * \sqrt{108} [/tex]
The first thing we are going to do is simplify the radicands 196 ans 108 (picture 1):
[tex]196=2^2*7^2[/tex] and [tex]108=2^2*3^3[/tex]
Knowing this, we can rewrite our radicals as follows:
[tex]\frac{ \sqrt{196} }{7} * \sqrt{108}= \frac{ \sqrt{2^2*7^2} }{7} * \sqrt{2^2*3^3} [/tex]
Remember that [tex] \sqrt[n]{x^n} =x[/tex]; in other words if the radicand is raised to the same power as the index of the radical, we can take the radicand out. Since 2 and 7 are raised to the power 2 and the index of the radical is also 2 (square root), we can take out 2 and 7:
[tex]\frac{ \sqrt{2^2*7^2} }{7} * \sqrt{2^2*3^3}= \frac{2*7}{7} *2 \sqrt{3^3} [/tex]
Look! we have the same numerator and denominator in our fraction, so we can cancel them both:
[tex]\frac{2*7}{7} *2 \sqrt{3^3}=2*2 \sqrt{3^3} =4 \sqrt{3^3} [/tex]
Notice that we can write [tex]3^3[/tex] as [tex]3^2*3[/tex], so we can rewrite our expression one last time:
[tex]4 \sqrt{3^3} =4 \sqrt{3^2*3} =4*3 \sqrt{3} =12 \sqrt{3} [/tex]
We can conclude that the correct option is: [tex]12 \sqrt{3} [/tex]
2. The product of a nonzero rational number and an irrational number is always an irrational number.
Proof by contradiction:
Lets assume that the product of an irrational number and a rational non-zero number is always rational.
Let [tex]x[/tex] be and irrational number and let [tex] \frac{a}{b} [/tex] and [tex] \frac{c}{d} [/tex] be two rational numbers with [tex]a[/tex], [tex]b[/tex], [tex]c[/tex], and [tex]d[/tex] are non-zero integers.
[tex]x* \frac{a}{b} = \frac{c}{d} [/tex]
[tex]x= \frac{c}{d} * \frac{b}{a} [/tex]
[tex]x=\frac{cb}{da}[/tex]
Since integers are closed under multiplication, [tex] \frac{cb}{da}[/tex] is a rational number. Since[tex]x[/tex] is an irrational number and [tex]x=\frac{cb}{da}[/tex], we have a logical contradiction, so we can conclude that the product of an irrational number and a rational non-zero number is always an irrational number.
