An irrational number between sqrt2 and sqrt3 is (a) (sqrt2+sqrt3) (b) sqrt2×sqrt3 (c) 5^(1//4) (d) 6^(1//4)

Question

An irrational number between  sqrt2 and  sqrt3 is (a)  (sqrt2+sqrt3) (b)  sqrt2×sqrt3 (c)  5^(1//4) (d)  6^(1//4)

Bytelearn · Accepted Answer

Calculate Approximate Square Roots: We need to find which option is an irrational number and lies between sqrt(2) and sqrt(3). Let's start by calculating the approximate values of sqrt(2) and sqrt(3). sqrt(2) ≈ 1.414, sqrt(3) ≈ 1.732.Check Option (a): Check option (a) sqrt(2) + sqrt(3). (sqrt(2) + sqrt(3)) ≈ 1.414 + 1.732 = 3.146. This value is greater than sqrt(3), so it's not between sqrt(2) and sqrt(3).Check Option (b): Check option (b) sqrt(2) × sqrt(3). (sqrt(2) × sqrt(3)) ≈ 1.414 × 1.732 = 2.449. This value is also greater than sqrt(3), so it's not between sqrt(2) and sqrt(3).Check Option (c): Check option (c) 5^(1//4). 5^(1//4) means the fourth root of 5, which is approximately 1.495. This value is greater than sqrt(2) but less than sqrt(3), so it could be a valid answer.Check Option (d): Check option (d) 6^(1//4). 6^(1//4) means the fourth root of 6, which is approximately 1.565. This value is also greater than sqrt(2) but less than sqrt(3), so it could also be a valid answer.

An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is $\newline$ (a) $(\sqrt{2}+\sqrt{3})$ $\newline$ (b) $\sqrt{2} \times \sqrt{3}$ $\newline$ (c) $5^{1 / 4}$ $\newline$ (d) $6^{1 / 4}$

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An irrational number between 2 \sqrt{2} 2​ and 3 \sqrt{3} 3​ is\newline(a) (2+3) (\sqrt{2}+\sqrt{3}) (2​+3​)\newline(b) 2×3 \sqrt{2} \times \sqrt{3} 2​×3​\newline(c) 51/4 5^{1 / 4} 51/4\newline(d) 61/4 6^{1 / 4} 61/4

An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is $\newline$ (a) $(\sqrt{2}+\sqrt{3})$ $\newline$ (b) $\sqrt{2} \times \sqrt{3}$ $\newline$ (c) $5^{1 / 4}$ $\newline$ (d) $6^{1 / 4}$