Prove square root of 3 is irrational by contradiction

Proof: Once again we will prove this by contradiction. Suppose that there exists a rational number $r = frac{a}{b}$ such that $r^2 = 3$.The number ?3 is irrational ,it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us Assume that it is rational and. To the contrary assume that sqrt(3) is a rational number. Thus we can write sqrt(3) = a/b where a and b are coprime integers and b is non-zero.Do you mean sqrt(3) – 5? Simple: assume that it is a rational number, r. Then so is sqrt(3) = 5 + r, contradiction, since it is very. Root 3 is equal to 1.732050807…. ?3 = 1.7320508…. where the numbers never terminate. This is an example of an irrational number, which is defined as a.

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