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Posted by Robin Pasrija 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
If possible,let √p be a rational number.
also a and b is rational.
then,√p = a/b
on squaring both sides,we get,
(√p)²= a²/b²
→p = a²/b²
→b² = a²/p [p divides a² so,p divides a]
Let a= pr for some integer r
→b² = (pr)²/p
→b² = p²r²/p
→b² = pr²
→r² = b²/p [p divides b² so, p divides b]
Thus p is a common factor of a and b.
But this is a contradiction, since a and b have no common factor.
This contradiction arises by assuming √p a rational number.
Hence,√p is irrational.
Posted by Robin Pasrija 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
Let us assume that 5 - √3 is rational. Then it can be written in the form
5 - √3 = p/q
or 5 - p/q = √3
It implies root3 is a rational number [Since 5 - p/q are rationals]
But this contradicts to the fact that √3 is irrational. Hence our supposition was wrong. Therefore 5 - √3 is irrational.
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Posted by Sëlfìê ?? 6 years, 2 months ago
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Shreya ... 6 years, 2 months ago
81=75×1+6
75=6×12+3
6=3×2+0 .....so, HCF will be 3.
3=75-6×12◆75-(81-75×1)×12◆75+75×12-81×12◆75(1+12)-81×12◆75×12-81×12◆(237-81×2)×13-81×2◆237×13-81×2×13-81×12◆237×13-81(26+12)◆237×13-81×38◆81×(-38)+237×13◆81x+237y...hence, x will be -38 and y will be 13
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