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Sagot :
The [tex]\sqrt{2}[/tex] is not a rational number, it is an irrational number
The number is [tex]\sqrt{2}[/tex]
The rational number is the number that can be written in the form of p/q where q ≠ 0
Therefore assume [tex]\sqrt{2}[/tex] is a rational number
[tex]\sqrt{2}[/tex] = p/q
Find the square of the both side
2 = p^2 / q^2
2q^2 = p^2
The value of p^2 will be the multiple of 2
p will be the multiple of 2
Consider the multiple as m
p = 2m
p^2 = 4m^2
Substitute the values in the above equation
2q^2 = 4m^2
q^2 = 2m^2
Therefore the value of the q will be the multiple of 2
The p and q is the multiple of 2, therefore they have coprime
Our assumption is wrong, [tex]\sqrt{2}[/tex] is an irrational number
Therefore, the [tex]\sqrt{2}[/tex] is not a rational number
Learn more about irrational number here
brainly.com/question/17450097
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