Question

Is √2 a rational number if we write it as √2/1​

Answer 1

no root 2 is not a rational number it is an irrational numberwhen we write root 2 by 1 it can also become an irrational number but not a rational number

Answer 2

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$\bold{\underline{\underline{QUESTION}}}$

IS ROOT 2 A RATIONAL NUMBER WHY OR WHY NOT ??

$\bold{\underline{\underline{ANSWER}}}$

NO, ROOT 2 IS NOT A RATIONAL NUMBER.

$\bold{\underline{\underline{Step\:To\:Step\:Explanation}}}$

Proof by contradiction

Assume √2 is a rational number. Goal: Show that this leads to a contradiction. If so, then √2 is not a rational number.If √2 is a rational number, then √2 = p⁄q for some p, q ∈ ℤ where q ≠ 0 (by the definition of ℚ)(✭) Further, we assume that p⁄q is in lowest terms.2)

Then, by algebra:

Then, by algebra:√2 = p⁄q

Then, by algebra:√2 = p⁄q2 = p2⁄q2

Then, by algebra:√2 = p⁄q2 = p2⁄q22q2 = p2

Then, by algebra:√2 = p⁄q2 = p2⁄q22q2 = p2If p2 = 2q2, then p2 is even (by the definition of even).

Then, by algebra:√2 = p⁄q2 = p2⁄q22q2 = p2If p2 = 2q2, then p2 is even (by the definition of even).If p2 is even, then p is even (by the proof above).

Then, by algebra:√2 = p⁄q2 = p2⁄q22q2 = p2If p2 = 2q2, then p2 is even (by the definition of even).If p2 is even, then p is even (by the proof above).If p is even, then p = 2r for some r ∈ ℤ (by the definition of even).

Then, by algebra:√2 = p⁄q2 = p2⁄q22q2 = p2If p2 = 2q2, then p2 is even (by the definition of even).If p2 is even, then p is even (by the proof above).If p is even, then p = 2r for some r ∈ ℤ (by the definition of even).Then by substitution: 2q2 = (2r)2

Then, by algebra:√2 = p⁄q2 = p2⁄q22q2 = p2If p2 = 2q2, then p2 is even (by the definition of even).If p2 is even, then p is even (by the proof above).If p is even, then p = 2r for some r ∈ ℤ (by the definition of even).Then by substitution: 2q2 = (2r)2By algebra:

2q2 = (2r)2

2q2 = (2r)22q2 = 4r2

2q2 = (2r)22q2 = 4r22q2 = 2(2r2)

2q2 = (2r)22q2 = 4r22q2 = 2(2r2)q2 = 2r2

But if q2 = 2r2, then q2 is even (by the definition of even numbers). And, if q2 is even, then q is even (by the argument above).

But if q2 = 2r2, then q2 is even (by the definition of even numbers). And, if q2 is even, then q is even (by the argument above).Thus, we have shown that both p and q are even. But if p and q are even, then they both have two as a factor (by the definition of even number. Thus, p and q are not in lowest terms. But this contradicts our earlier assumption (✭), and thus the √2 is not a rational number.

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