What numbers added to #0.4# will produce an irrational number?

2 Answers
Mar 22, 2017

You didn't provide what numbers to choose, but any irrational number plus a rational number produces another irrational number. For instance...

#pi + 0.4# is irrational. You get something like #3.5415926535cdots#

#sqrt(2) + 0.4# is irrational. You get something like #1.814213cdots#

So the number with infinitely repeating decimals (the one that looks like it will have more digits than your calculator wants to show) is irrational. By definition it means it cannot be represented as a fraction with whole numbers in the top and bottom of the fraction.

So:

#35/50# is not irrational
#1.2434# is not irrational if this is exact
#sqrt3# is irrational because it cannot be represented as a fraction with all whole numbers in the top and bottom.
#pi (3.1415926535cdots)# and #e(2.718281828cdots)# are irrational
etc (see this article to see how #e# is irrational).

Mar 22, 2017

Answer:

Any irrational number when added to 0.4 produce an irrational sum.

Example: #0.4+pi# is irrational

Explanation:

There are irrational numbers that are special and crop up all over the place. One of them is used in working out the area of a circle and that number is #pi#. As far as they know the decimal digits just go on for ever and does not have a repeating cycle

Add a rational number such as #0.4->4/10# to an irrational number and the decimal digits of the sum will also go on forever without a continual repeat.

So #0.4+pi# is irrational
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A rational number can be expressed as a fraction. That is, it has the form #a/b# where #b!=0#. They have a decimal equivalent that terminates ( stops) or has a cycle of repeated digits.

A well known on is #1/3#. This has a repeat #->0.3333333....#
A denominator of 11 is a good one #->3/11 = 0.27272727....#

An irrational number can #ul("not")# be written in the form of #a/b#

Note that #a and b# are whole numbers.