Prove that `\sqrt{7}` is irrational.
Let us assume, to the contrary, that ,√7 is a rational number. Then, there exist co-prime positive integers a and b such that
`\sqrt{7}` = `"a"/"b"` b ≠ 0
So, a `\sqrt{7}`b .....(i)
Squaring both sides, we have
`"a"^2` = 7`"b"^2` = 7 divides a
So we can write
a = 7c, (where c is any integer)
Putting the value of a = 7 c in (i), we have
49 `"c"^2` = 7`"b"^2`
7 `"c"^2` = `"b"^2`
It means 7 divides `"b"^2` and so 7 divides b.
So, 7 is a common factor of both a and b which is a contradiction.
So, our assumption that `\sqrt{7}` is a rational is wrong.
Hence, we conclude that `\sqrt{7}` is a irational number.
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