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Sagot :
To prove that [tex]\(\sqrt{7}\)[/tex] is an irrational number, we can use a proof by contradiction. Here's how we do it step-by-step:
1. Assume the Opposite:
Let's assume that [tex]\(\sqrt{7}\)[/tex] is a rational number. By definition, a rational number can be expressed as a fraction [tex]\(\frac{a}{b}\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers with no common divisors other than 1 (i.e., in simplest form) and [tex]\(b \neq 0\)[/tex].
2. Expressing the Assumption:
Based on our assumption, we have:
[tex]\[ \sqrt{7} = \frac{a}{b} \][/tex]
3. Square Both Sides:
To eliminate the square root, we square both sides of the equation:
[tex]\[ (\sqrt{7})^2 = \left(\frac{a}{b}\right)^2 \][/tex]
This simplifies to:
[tex]\[ 7 = \frac{a^2}{b^2} \][/tex]
4. Clear the Fraction:
Multiply both sides of the equation by [tex]\(b^2\)[/tex] to get rid of the denominator:
[tex]\[ 7b^2 = a^2 \][/tex]
5. Analyze the Equation:
This implies that [tex]\(a^2\)[/tex] is a multiple of 7. In other words, [tex]\(a^2\)[/tex] can be written as [tex]\(7k\)[/tex] for some integer [tex]\(k\)[/tex]. Because 7 is a prime number, it must be that [tex]\(a\)[/tex] itself is a multiple of 7. Let's denote this by:
[tex]\[ a = 7m \quad \text{for some integer } m \][/tex]
6. Substitute Back:
Substitute [tex]\(a = 7m\)[/tex] into the equation [tex]\(7b^2 = a^2\)[/tex]:
[tex]\[ 7b^2 = (7m)^2 \][/tex]
This simplifies to:
[tex]\[ 7b^2 = 49m^2 \][/tex]
Dividing both sides by 7 results in:
[tex]\[ b^2 = 7m^2 \][/tex]
7. Analyze the New Equation:
This implies that [tex]\(b^2\)[/tex] is also a multiple of 7. Using similar reasoning as before, because 7 is a prime number, [tex]\(b\)[/tex] must also be a multiple of 7. Suppose:
[tex]\[ b = 7n \quad \text{for some integer } n \][/tex]
8. Contradiction:
We have shown that both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are multiples of 7. Hence, [tex]\(a=7m\)[/tex] and [tex]\(b=7n\)[/tex]. However, this leads to a contradiction because we initially assumed that [tex]\( \frac{a}{b} \)[/tex] is in simplest form (i.e., [tex]\(a\)[/tex] and [tex]\(b\)[/tex] have no common divisors other than 1). If both are multiples of 7, then they have 7 as a common divisor, contradicting our assumption.
9. Conclusion:
Since our assumption that [tex]\(\sqrt{7}\)[/tex] is a rational number leads to a contradiction, we conclude that [tex]\(\sqrt{7}\)[/tex] must be irrational.
Therefore, [tex]\(\sqrt{7}\)[/tex] is an irrational number.
1. Assume the Opposite:
Let's assume that [tex]\(\sqrt{7}\)[/tex] is a rational number. By definition, a rational number can be expressed as a fraction [tex]\(\frac{a}{b}\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers with no common divisors other than 1 (i.e., in simplest form) and [tex]\(b \neq 0\)[/tex].
2. Expressing the Assumption:
Based on our assumption, we have:
[tex]\[ \sqrt{7} = \frac{a}{b} \][/tex]
3. Square Both Sides:
To eliminate the square root, we square both sides of the equation:
[tex]\[ (\sqrt{7})^2 = \left(\frac{a}{b}\right)^2 \][/tex]
This simplifies to:
[tex]\[ 7 = \frac{a^2}{b^2} \][/tex]
4. Clear the Fraction:
Multiply both sides of the equation by [tex]\(b^2\)[/tex] to get rid of the denominator:
[tex]\[ 7b^2 = a^2 \][/tex]
5. Analyze the Equation:
This implies that [tex]\(a^2\)[/tex] is a multiple of 7. In other words, [tex]\(a^2\)[/tex] can be written as [tex]\(7k\)[/tex] for some integer [tex]\(k\)[/tex]. Because 7 is a prime number, it must be that [tex]\(a\)[/tex] itself is a multiple of 7. Let's denote this by:
[tex]\[ a = 7m \quad \text{for some integer } m \][/tex]
6. Substitute Back:
Substitute [tex]\(a = 7m\)[/tex] into the equation [tex]\(7b^2 = a^2\)[/tex]:
[tex]\[ 7b^2 = (7m)^2 \][/tex]
This simplifies to:
[tex]\[ 7b^2 = 49m^2 \][/tex]
Dividing both sides by 7 results in:
[tex]\[ b^2 = 7m^2 \][/tex]
7. Analyze the New Equation:
This implies that [tex]\(b^2\)[/tex] is also a multiple of 7. Using similar reasoning as before, because 7 is a prime number, [tex]\(b\)[/tex] must also be a multiple of 7. Suppose:
[tex]\[ b = 7n \quad \text{for some integer } n \][/tex]
8. Contradiction:
We have shown that both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are multiples of 7. Hence, [tex]\(a=7m\)[/tex] and [tex]\(b=7n\)[/tex]. However, this leads to a contradiction because we initially assumed that [tex]\( \frac{a}{b} \)[/tex] is in simplest form (i.e., [tex]\(a\)[/tex] and [tex]\(b\)[/tex] have no common divisors other than 1). If both are multiples of 7, then they have 7 as a common divisor, contradicting our assumption.
9. Conclusion:
Since our assumption that [tex]\(\sqrt{7}\)[/tex] is a rational number leads to a contradiction, we conclude that [tex]\(\sqrt{7}\)[/tex] must be irrational.
Therefore, [tex]\(\sqrt{7}\)[/tex] is an irrational number.
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