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Ankesh ..?? Now I Am Fully Ready? 6 years ago
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Yogita Ingle 6 years ago
Let us assume that √3 is a rational number.
then, as we know a rational number should be in the form of p/q
where p and q are co- prime number.
So, √3 = p/q { where p and q are co- prime}
√3q = p
Now, by squaring both the side
we get,
(√3q)² = p²
3q² = p² ........ ( i )
So, if 3 is the factor of p²
then, 3 is also a factor of p ..... ( ii )
=> Let p = 3m { where m is any integer }
squaring both sides
p² = (3m)²
p² = 9m²
putting the value of p² in equation ( i )
3q² = p²
3q² = 9m²
q² = 3m²
So,
if 3 is factor of q²
then, 3 is also factor of q
Since 3 is factor of p & q both
So, our assumption that p & q are co- prime is wrong
Hence,. √3 is an irrational number.
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Posted by Secret Star🌠🌠🌠🌠 6 years ago
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Aadya Singh ? 6 years ago
=> -2 = 10+(3-1)×(k-10)
=> -2 = 10+2k-20
=> -2 +10 = 2k
=>k =8/2 => k = 4.
Posted by Anas Kassar 6 years ago
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Secret Star🌠🌠🌠🌠 6 years ago
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Aadya Singh ? 6 years ago
4Thank You