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To prove that [tex]\( \sqrt{\frac{1}{5}} \)[/tex] is an irrational number, let's follow a series of logical and mathematical steps to demonstrate the proof effectively. Note that a number is considered rational if it can be expressed as the ratio of two integers (i.e., in the form [tex]\(\frac{a}{b}\)[/tex] where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers and [tex]\(b \neq 0\)[/tex], and they have no common factors other than 1). If such a form cannot be found, the number is irrational.
1. Assume [tex]\( \sqrt{\frac{1}{5}} \)[/tex] is rational:
Suppose [tex]\( \sqrt{\frac{1}{5}} \)[/tex] is rational. Then, it can be written as [tex]\(\frac{a}{b}\)[/tex] where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers, and [tex]\(b \neq 0\)[/tex], and [tex]\(\frac{a}{b}\)[/tex] is in its simplest form (i.e., [tex]\(a\)[/tex] and [tex]\(b\)[/tex] have no common factors other than 1).
2. Express the equality in terms of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ \sqrt{\frac{1}{5}} = \frac{a}{b} \][/tex]
3. Square both sides:
[tex]\[ \left( \sqrt{\frac{1}{5}} \right)^2 = \left( \frac{a}{b} \right)^2 \][/tex]
This simplifies to:
[tex]\[ \frac{1}{5} = \frac{a^2}{b^2} \][/tex]
4. Cross-multiply to clear the fraction:
[tex]\[ 1 \cdot b^2 = 5 \cdot a^2 \][/tex]
Therefore:
[tex]\[ b^2 = 5a^2 \][/tex]
5. Analyze the implications of the equation [tex]\( b^2 = 5a^2 \)[/tex]:
- The left-hand side, [tex]\(b^2\)[/tex], is a perfect square.
- For [tex]\(b^2\)[/tex] to be a product involving 5, [tex]\(a^2\)[/tex] must also be a perfect square and must be multiplied by 5.
- This implies that [tex]\(b^2\)[/tex] must be a multiple of 5.
6. Conclude the form of [tex]\(b\)[/tex]:
Since [tex]\(b^2\)[/tex] is a multiple of 5, [tex]\(b\)[/tex] must itself be a multiple of 5 (because the square of any number that is not a multiple of 5 cannot result in a multiple of 5).
7. Let [tex]\(b = 5k\)[/tex] for some integer [tex]\(k\)[/tex]:
If [tex]\(b\)[/tex] is a multiple of 5, substitute [tex]\(b = 5k\)[/tex] into the equation found:
[tex]\[ (5k)^2 = 5a^2 \][/tex]
Simplify this:
[tex]\[ 25k^2 = 5a^2 \][/tex]
Divide both sides by 5:
[tex]\[ 5k^2 = a^2 \][/tex]
8. Conclude the form of [tex]\(a\)[/tex]:
This implies that [tex]\(a^2\)[/tex] is also a multiple of 5, meaning [tex]\(a\)[/tex] must be a multiple of 5 much like our reasoning for [tex]\(b\)[/tex].
9. Contradiction due to common factors:
Since both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are multiples of 5, they have a common factor of 5, contradicting our initial condition that [tex]\(a\)[/tex] and [tex]\(b\)[/tex] have no common factors other than 1.
10. Conclusion:
Since our initial assumption that [tex]\(\sqrt{\frac{1}{5}} = \frac{a}{b}\)[/tex] (in its simplest form) led to a contradiction, we conclude that [tex]\(\sqrt{\frac{1}{5}}\)[/tex] cannot be expressed as a ratio of two integers. Therefore, [tex]\(\sqrt{\frac{1}{5}}\)[/tex] is irrational.
Additionally, the numerical result of [tex]\(\sqrt{\frac{1}{5}}\)[/tex] is [tex]\(0.4472135954999579\)[/tex], and this further supports our conclusion that [tex]\(\sqrt{\frac{1}{5}}\)[/tex] is an irrational number.
1. Assume [tex]\( \sqrt{\frac{1}{5}} \)[/tex] is rational:
Suppose [tex]\( \sqrt{\frac{1}{5}} \)[/tex] is rational. Then, it can be written as [tex]\(\frac{a}{b}\)[/tex] where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers, and [tex]\(b \neq 0\)[/tex], and [tex]\(\frac{a}{b}\)[/tex] is in its simplest form (i.e., [tex]\(a\)[/tex] and [tex]\(b\)[/tex] have no common factors other than 1).
2. Express the equality in terms of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ \sqrt{\frac{1}{5}} = \frac{a}{b} \][/tex]
3. Square both sides:
[tex]\[ \left( \sqrt{\frac{1}{5}} \right)^2 = \left( \frac{a}{b} \right)^2 \][/tex]
This simplifies to:
[tex]\[ \frac{1}{5} = \frac{a^2}{b^2} \][/tex]
4. Cross-multiply to clear the fraction:
[tex]\[ 1 \cdot b^2 = 5 \cdot a^2 \][/tex]
Therefore:
[tex]\[ b^2 = 5a^2 \][/tex]
5. Analyze the implications of the equation [tex]\( b^2 = 5a^2 \)[/tex]:
- The left-hand side, [tex]\(b^2\)[/tex], is a perfect square.
- For [tex]\(b^2\)[/tex] to be a product involving 5, [tex]\(a^2\)[/tex] must also be a perfect square and must be multiplied by 5.
- This implies that [tex]\(b^2\)[/tex] must be a multiple of 5.
6. Conclude the form of [tex]\(b\)[/tex]:
Since [tex]\(b^2\)[/tex] is a multiple of 5, [tex]\(b\)[/tex] must itself be a multiple of 5 (because the square of any number that is not a multiple of 5 cannot result in a multiple of 5).
7. Let [tex]\(b = 5k\)[/tex] for some integer [tex]\(k\)[/tex]:
If [tex]\(b\)[/tex] is a multiple of 5, substitute [tex]\(b = 5k\)[/tex] into the equation found:
[tex]\[ (5k)^2 = 5a^2 \][/tex]
Simplify this:
[tex]\[ 25k^2 = 5a^2 \][/tex]
Divide both sides by 5:
[tex]\[ 5k^2 = a^2 \][/tex]
8. Conclude the form of [tex]\(a\)[/tex]:
This implies that [tex]\(a^2\)[/tex] is also a multiple of 5, meaning [tex]\(a\)[/tex] must be a multiple of 5 much like our reasoning for [tex]\(b\)[/tex].
9. Contradiction due to common factors:
Since both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are multiples of 5, they have a common factor of 5, contradicting our initial condition that [tex]\(a\)[/tex] and [tex]\(b\)[/tex] have no common factors other than 1.
10. Conclusion:
Since our initial assumption that [tex]\(\sqrt{\frac{1}{5}} = \frac{a}{b}\)[/tex] (in its simplest form) led to a contradiction, we conclude that [tex]\(\sqrt{\frac{1}{5}}\)[/tex] cannot be expressed as a ratio of two integers. Therefore, [tex]\(\sqrt{\frac{1}{5}}\)[/tex] is irrational.
Additionally, the numerical result of [tex]\(\sqrt{\frac{1}{5}}\)[/tex] is [tex]\(0.4472135954999579\)[/tex], and this further supports our conclusion that [tex]\(\sqrt{\frac{1}{5}}\)[/tex] is an irrational number.
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