C-01-Q037mediumsingle_mcq

In the proof that $\sqrt{2}$ is irrational, $\sqrt{2}$ is initially assumed as—

aa natural number
ba rational number $\frac{p}{q}$ where $p, q$ are mutually prime and $q > 1$
can integer
dan imaginary number

ব্যাখ্যা

The proof uses contradiction: it begins by assuming $\sqrt{2}$ is rational, written as $\frac{p}{q}$ in lowest terms with $p, q$ mutually prime. This assumption later leads to a contradiction, proving $\sqrt{2}$ is irrational.

এই অধ্যায়ের আরও প্রশ্ন

C-01-Q038Which of the following is NOT an irrational number?
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C-01-Q040Which of the following is a finite decimal fraction?
C-01-Q041is a—
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C-01-Q043Finite and repeating decimal fractions are—
C-01-Q044Infinite non-repeating decimal numbers are—
C-01-Q045means—
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Real Numbers — সব প্রশ্ন দেখুন (149টি) →

In the proof that $\sqrt{2}$ is irrational, $\sqrt{2}$ is initially assumed as—

aa natural number

ba rational number

\frac{p}{q}

where

p, q

are mutually prime and

q > 1

can integer

dan imaginary number

In the proof that 2\sqrt{2}2​ is irrational, 2\sqrt{2}2​ is initially assumed as—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof that 2\sqrt{2}2​ is irrational, 2\sqrt{2}2​ is initially assumed as—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In the proof that $\sqrt{2}$ is irrational, $\sqrt{2}$ is initially assumed as—

In the proof that $\sqrt{2}$ is irrational, $\sqrt{2}$ is initially assumed as—