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Mathematics 19 Online
OpenStudy (anonymous):

identify the number that does not belong with the other three. Explain your reasoning. @mathstudent55

OpenStudy (anonymous):

OpenStudy (anonymous):

i think it 13.4

OpenStudy (mathstudent55):

I think this question deals with sets of numbers. Have you learned the sets of numbers called: Natural Numbers Whole Numbers Rational Numbers Irrational Numbers Real Numbers

OpenStudy (anonymous):

its hard for me to understand math.

OpenStudy (mathstudent55):

Natural numbers: 1, 2, 3, 4, 5, ... Whole numbers: 0, 1, 2, 3, 4, 5, ... Integers: ..., -3, -2, -1, 0, 1, 2, 3, ... Rational Number: Any number that can be written as a fraction of integers Irrational Number: a number that is not rational Real Numbers: the set containing all rational and irrational numbers

OpenStudy (mathstudent55):

Look at each number and see which sets it belongs to.

OpenStudy (anonymous):

okay 22.7 belongs to rational numbers? -13.4 belongs to integers? -10/2 belongs to rational? 22.7 belongs to integers.

OpenStudy (mathstudent55):

-10/2 = -5 -5 is an integer and is rational

OpenStudy (mathstudent55):

-13.4 is not an integer, but it is rational. It can be written as -134/10, for example.

OpenStudy (anonymous):

oh yeah! oh my! i forgot you can add a 0 to any 1 with rational numbers.

OpenStudy (mathstudent55):

22.777... (the bar over the seven means the seven repeats). This is a rational number. Any repeating decimal can be written as a fraction of integers.

OpenStudy (anonymous):

i think its 18 that doesnt belong there.

OpenStudy (mathstudent55):

That leaves \(\sqrt{18} \). This can be simplified like this: \(\sqrt{18} = \sqrt{9 \times2} = \sqrt 9 \times \sqrt 2 = 3\sqrt{2}\) The square root of 2 is not a rational number, so \(\sqrt{18} \) is an irrational number.

OpenStudy (mathstudent55):

You are right. All other numbers are rational. \(\sqrt{18} \) is the only irrational number in the group.

OpenStudy (anonymous):

yes! you made me understand! Thank you very much! Do you mind helping me with 4 more questions? @mathstudent55

OpenStudy (mathstudent55):

Ok, but I need to go in about 5 minutes. Start a new post with one, and we'll do it together.

OpenStudy (anonymous):

yay! okay! @mathstudent55

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