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Mathematics
Kymani Mccullough
11 March, 04:33
Prove that 7 + 3 root 2 is not a rational number
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Dharma
11 March, 06:06
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Prove by contradiction.
See below.
Step-by-step explanation:
Assume it is rational, then:
Let x = 7 + 3√2 = p/q where p and q are integers and p/q is in simplest form.
x^2 = 49 + 42√2 + 18 = p^2/q^2
67q^2 = p^2 - 42√2q^2
67q^2 - p^2 = - 42√2 q^2
67q^2 - p^2 is rational and q^2 is rational so that makes - 42√2 rational.
but we know that - 42√2 is irrational, so our original assumption is not true.
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Adalynn Simpson
11 March, 06:25
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Step-by-step explanation:
let 7 + 3√2 be an rational number where
7+3√2 = a/b [ a and b are coprime and b is not equal to zero]
3√2 = a/b-7
3√2 = (a-7b) / b
√2 = (a-7b) / 3b ... (i)
Now, from equation (i), we get that √2 is rational but we know that √2 is irrational. so actually 7 + 3√2 is irrational not rational. thus our assumption is wrong. The number is irrational.
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