[tex]prove \: that \: \sqrt{3 \: } \: \: is \: an \: irrational \: no. \: .[/tex]
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[tex]prove \: that \: \sqrt{3 \: } \: \: is \: an \: irrational \: no. \: .[/tex]
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Step-by-step explanation:
Say 3–√ is rational.
Then 3–√ can be represented as ab, where a and b have no common factors.
[tex]3 = \frac{a {}^{2} }{b {}^{2} } [/tex]
[tex]3b {}^{2} = a {}^{2} [/tex]
Now a2 must be divisible by 3, but then so must a (fundamental theorem of arithmetic).
[tex]3b {}^{2} =( 3k) {}^{2} [/tex]
[tex]b {}^{2} = 3k {}^{2} [/tex]
and now we have a contradiction
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Verified answer
Answer:
Co prime:
Two numbers are called coprime or relative primes, if they have only one common factor which is 1 e.g., (7,9) is coprime but 15 and 21 is not coprime.
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