The rational numbers are dense in RR. This means that between any two real numbers a and b with a < b, there exists a rational number q such that a < q < b. Using this fact, establish that the irrational numbers are dense in RR as well.

The rational numbers are dense in RR. This means that between any two real numbers a and b with a < b, there exists a rational number q such that a < q < b. Using this fact, establish that the irrational numbers are dense in RR as well.

Question
Irrational numbers
asked 2021-01-23
The rational numbers are dense in \(\displaystyle\mathbb{R}\). This means that between any two real numbers a and b with a < b, there exists a rational number q such that a < q < b. Using this fact, establish that the irrational numbers are dense in \(\displaystyle\mathbb{R}\) as well.

Answers (1)

2021-01-24
We need to prove that Irrational numbers are dense in \(\displaystyle\mathbb{R}\) .We are given that Rational numbers are dense in \(\displaystyle\mathbb{R}\) , which means between any two real numbers there exist a rational number. Now, \(\displaystyle\sqrt{{2}}\) is an irrational number .
Let, a and b be two real numbers, such that, a Since \(\displaystyle{a},{b},\sqrt{{2}}\) are real numbers therefore there division is also a real number
Therefore , \(\displaystyle\frac{{a}}{\sqrt{{2}}}\) and b/sqrt2ZSK are also real numbers. We have,
\(\displaystyle{a}{<}{b}\Rightarrow\frac{{a}}{\sqrt{{2}}}{<}\frac{{b}}{\sqrt{{2}}}\)</span>
And we are given that between any two real numbers there exist a rational number .
Therefore , let c in \(\displaystyle\mathbb{Q}\) be a rational number between \(\displaystyle\frac{{a}}{\sqrt{{2}}}\) and \(\displaystyle\frac{{b}}{\sqrt{{2}}}\Rightarrow\frac{{a}}{\sqrt{{2}}}{<}{c}{<}\frac{{b}}{\sqrt{{2}}}\)</span>
We got,
\(\displaystyle\frac{{a}}{\sqrt{{2}}}{<}{c}{<}\frac{{b}}{\sqrt{{2}}}\)</span>
Now , multiply by \(\displaystyle\sqrt{{2}}\) throughout, we get,
\(\displaystyle\sqrt{{2}}{\left(\frac{{a}}{\sqrt{{2}}}\right)}{<}\sqrt{{2}}\cdot{c}{<}\sqrt{{2}}{\left(\frac{{b}}{\sqrt{{2}}}\right)}\Rightarrow{a}{<}{c}\sqrt{{2}}{<}{b}\)</span>
Now, product of a rational and an irrational number is an irrational number , therefore , since c in \(\displaystyle\mathbb{Q}\) and \(\displaystyle\sqrt{{2}}\in\frac{\mathbb{R}}{\mathbb{Q}}\), therefore, csqrt2ZSK
Let, \(\displaystyle{c}\sqrt{{2}}={m}\Rightarrow{a}{<}{m}{<}{b}\)</span>
and a and b are real numbers and m is irrational number .
Therefore, we get that between any two real numbers a and b, with a Therefore, Irrational numbers are dense in real numbers .
Answer: Irrational numbers are dense in \(\displaystyle\mathbb{R}\) .
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