[tex] \sqrt{5} [/tex]
prove this is irrational no................❤️❤️❤️
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[tex] \sqrt{5} [/tex]
prove this is irrational no................❤️❤️❤️
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Let's take [tex] \sqrt{5} [/tex] as rational number .
We can write [tex] \sqrt{5} = \frac{a}{b} [/tex] such that ,
a & b are integers , [tex] b ≠ 0 [/tex] & there are no factors common to a & b .
Multiplying b on both sides , we get ;
[tex] b \sqrt{5} = a [/tex]
To remove root , squaring on both sides , we get , [tex] 5b² = a² [/tex] _____ ( i )
That means , 5 is a factor of a² .
For any prime no. p which is a factor a² then it will be the factor of a also .
So , 5 is a factor of a . _____ ( ii )
Hence , we can write a = 5c for some integer c .
Putting the value of a in equation ( i ) we get ,
[tex] 5b² = (5c)² [/tex]
[tex] 5b² = 25c² [/tex]
Divide by 5 ; we get ,
[tex] b² = 5c² [/tex]
It means 5 is a factor of b²
Thus , 5 is a factor of b _____ ( iii )
From ( ii ) & ( iii ) , we can say that 5 is the factor of both a & b .
This contradicts our theory , as we stated that a & b have no factors in common
Thus , our assumption that [tex] \sqrt{5} [/tex] is rational is wrong .
Hence , [tex] \sqrt{5} [/tex] is irrational number .
Answer:
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