Ms. Fox asked her class is the sum of 4.2 and square root of 2 rational or irrational? Patrick answered that the sum would be irrational. State whether Patrick is correct or incorrect. Justify your reasoning.

2 Answers
May 17, 2018

The sum #4.2 + sqrt2# is irrational; it inherits the never-repeating decimal expansion property of #sqrt 2#.

Explanation:

An irrational number is a number that cannot be expressed as a ratio of two integers. If a number is irrational, then its decimal expansion goes on forever without a pattern, and vice versa.

We already know that #sqrt 2# is irrational. Its decimal expansion begins:

#sqrt 2 = 1.414213562373095...#

The number #4.2# is rational; it can be expressed as #42/10.# When we add 4.2 to the decimal expansion of #sqrt 2#, we get:

#sqrt 2 + 4.2 = color(white)+ 1.414213562373095...#
#color(white)(sqrt 2) color(white)+ color(white)(4.2 = )+ 4.2#
#color(white)(sqrt 2) color(white)+ color(white)(4.2 = )bar(color(white)(+)5.614213562373095...)#

It is easily seen that this sum also does not terminate nor have a repeating pattern, so it is also irrational.

In general, the sum of a rational number and an irrational number will always be irrational; the argument is similar to above.

May 17, 2018

#color(blue)("correct")#

Explanation:

If we start by saying the sum is rational: All rational numbers can be written as the quotient of two integers #a/bcolor(white)(88)# #b!=0#

#4.2=21/5#

#21/5+sqrt(2)=a/b#

#sqrt(2)=a/b-21/5#

#sqrt(2)=(5a-21b)/(5b)#

The product of two integers is an integer:

The difference of two integers is an integer:

So:

#5a-21b# is an integer.

#5b# is an integer.

Hence:

#(5a-21b)/(5b)# is rational.

But we know that #sqrt(2)# is irrational, so this is a contradiction from our assumption that the sum was rational, therefore the sum of an irrational number and a rational number is always irrational.