What is the square root of 13
1 answer:
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1. From your description, I can infer that the multiplication is:
The first thing we are going to do is simplify the radicands 196 ans 108 (picture 1):
and 
Knowing this, we can rewrite our radicals as follows:
Remember that![\sqrt[n]{x^n} =x](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bx%5En%7D%20%3Dx)
; 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:
Look! we have the same numerator and denominator in our fraction, so we can cancel them both:
Notice that we can write
as 
, so we can rewrite our expression one last time:
We can conclude that the correct option is:
2. The <span>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 </span>
be and irrational number and let 
and 
be two rational numbers with 
, 
, 
, and 
are non-zero integers.
Since integers are closed under multiplication,
is a rational number. Since is an irrational number and , 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.
The first thing we are going to do is simplify the radicands 196 ans 108 (picture 1):
Knowing this, we can rewrite our radicals as follows:
Remember that
Look! we have the same numerator and denominator in our fraction, so we can cancel them both:
Notice that we can write
We can conclude that the correct option is:
2. The <span>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 </span>
Since integers are closed under multiplication,