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Hard · Level 16 · sqrt2 proof,lowest form role,hard,class 10View options
To show contradiction when both (p) and (q) are later found even
To find decimal value
To make the denominator zero
To prove (\sqrt{2}=2)
Hard · Level 16 · sqrt5 proof,algebra error,hard,class 10View options
Incorrectly taking a root-level equation from a squared equation
Correctly writing the coprime condition
Correct use of prime factor
Correct identification of lowest form
Hard · Level 16 · sqrt3 proof,pre final step,hard,class 10View options
Both (p) and (q) are divisible by (3), which contradicts being coprime
(q=0), so the fraction cannot be formed
(\sqrt{3}=3), so it is a contradiction
(p=q), so the proof is complete
Hard · Level 16 · squaring role,irrationality proof,hard,class 10View options
Squaring removes the radical and makes reasoning about prime factor divisibility possible
Squaring makes the denominator zero
Squaring makes every number a perfect square
Squaring automatically proves rationality
Hard · Level 16 · sqrt5 proof,lowest form contradiction,hardView options
This result is impossible because the fraction can be reduced
This result proves (\sqrt{5}=5)
This result shows (q=0)
This result shows (5) is a perfect square
Hard · Level 16 · sqrt2 proof,gcd contradiction,hard,class 10View options
(\gcd(p,q)=1)
(q\neq 0)
(p) is an integer
(\sqrt{2}) is positive
Hard · Level 16 · sqrt3 proof,substitution reasoning,hardView options
(9k^2=3q^2), so (q^2=3k^2), hence (q) is divisible by (3)
(3k^2=3q^2), so (q=k), hence proof complete
(p=q), so contradiction
(q=0), so fraction invalid
Hard · Level 16 · sqrt5 misconception,rational square root,hardView options
The square root of a rational number is necessarily rational only when it is in a suitable perfect-square form
The square root of every rational number is an integer
(5) is not rational
(\sqrt{5}=5)
Hard · Level 16 · final conclusion,exam writing,hard,class 10View options
This contradicts our rational assumption, hence the given number is irrational
Therefore the given number is a perfect square
Therefore the denominator is zero
Therefore the square root equals the number inside
Hard · Level 16 · proof structure,irrationality,real numbers,hardView options
Assume rational, write a lowest-form fraction, square, get contradiction from a common factor
Find decimal, estimate, write the answer
Treat the square root as the number inside, then compare
Assume denominator zero, then remove the fraction
Hard · Level 17 · sqrt2 proof,even square,hard,class 10View options
(a^2) is even
(b) is even
(a=b)
(b=0)
Hard · Level 17 · sqrt3 proof,prime divisibility,hard,class 10View options
(p) is divisible by (2)
(p) is divisible by (3)
(p=q)
(p=1)
Hard · Level 17 · sqrt5 proof,substitution,hard,class 10View options
To show (m=n)
To show (n) is also divisible by (5)
To show (\sqrt{5}=5)
To show (n=0)
Hard · Level 17 · sqrt2 proof,b square,hard,class 10View options
(b^2=4r^2)
(b^2=r^2)
(b^2=2r^2)
(b=2r)
Hard · Level 17 · sqrt3 proof,algebra,hard,class 10View options
(9k^2=3q^2)
(3k^2=3q^2)
(k^2=3q^2)
(p^2=9q^2)
Hard · Level 17 · sqrt5 proof,algebra error,hard,class 10View options
From (p=5k), (p^2=25k^2)
From (p^2=5q^2), (p^2) is divisible by (5)
From (p=5k), (p^2=5k^2)
From (q^2=5k^2), (q) is divisible by (5)
Hard · Level 17 · coprime,common factor,hard,class 10View options
Both (a) and (b) are integers
Both (a) and (b) are divisible by (3)
(b\neq 0)
(a) is positive
Hard · Level 17 · sqrt2 proof,logical order,hard,class 10View options
Because to prove (b) even, (a=2k) must be substituted in the equation
Because (b) can never be even
Because (b=0)
Because (a=b) is already proved
Hard · Level 17 · sqrt3 proof,gcd,hard,class 10View options
(\gcd(p,q)) is at least (3)
(\gcd(p,q)) will remain (1)
(\gcd(p,q)=0)
(\gcd(p,q)=p+q)
Hard · Level 17 · sqrt5 proof,immediate conclusion,hard,class 10View options
(p^2) is divisible by (5)
(p) is divisible by (5)
(p=5k) for some integer (k)
(q) is divisible by (5)
Question 1HardLevel 16
Which option tells the role of assuming (\frac{p}{q}) in lowest form in the proof of (\sqrt{2})?
Correct answer: A
Step 1: In lowest form, (p) and (q) are coprime. Step 2: The proof shows both (p) and (q) are even. Step 3: Both being even breaks the lowest-form condition.
If someone writes (p=5q) from (p^2=5q^2) in the proof of (\sqrt{5}), what type of error is it?
Correct answer: A
Step 1: (p=5q) does not directly follow from (p^2=5q^2). Step 2: The correct conclusion is that (p^2) is divisible by (5), then (p) is divisible by (5). Step 3: Do not hastily derive a root-level equation from a squared equation.
In the proof of irrationality of (\sqrt{3}), which statement should come just before the final conclusion?
Correct answer: A
Step 1: The proof shows both (p) and (q) are divisible by (3). Step 2: This contradicts their coprime condition. Step 3: After this, the final conclusion is written that (\sqrt{3}) is irrational.
Which statement deeply explains the role of squaring in the proofs of (\sqrt{2}), (\sqrt{3}), and (\sqrt{5})?
Correct answer: A
Step 1: Squaring (\sqrt{n}) gives (n). Step 2: This forms an equation like (p^2=nq^2), which provides the base for divisibility. Step 3: Without this step, it is hard to create the common-factor contradiction.
In the proof of (\sqrt{5}), if (\frac{p}{q}) is in lowest form, which statement is correct when (p=5k) and (q=5r) are obtained?
Correct answer: A
Step 1: (p=5k) and (q=5r) mean both have common factor (5). Step 2: Such a fraction can be reduced by (5). Step 3: Hence this is an impossible result for the lowest-form assumption.
If both (p) and (q) are found even in the proof of (\sqrt{2}), which statement does it refute?
Correct answer: A
Step 1: If both are even, both (p) and (q) are divisible by (2). Step 2: Then their greatest common divisor cannot remain (1). Step 3: Therefore the condition (\gcd(p,q)=1) is refuted.
Which statement proves that just because (5) is rational, (\sqrt{5}) does not become rational?
Correct answer: A
Step 1: (5) is rational, but it is not a perfect square. Step 2: If it is not a perfect square, its square root need not be rational. Step 3: The proof of (\sqrt{5}) shows it is actually irrational.
Which option is the safest way to write the final conclusion in all three proofs?
Correct answer: A
Step 1: All three proofs begin with the rational assumption. Step 2: At the end, a common factor contradicts the coprime condition. Step 3: Therefore the final line should clearly state both contradiction and irrationality.
Which option best describes the correct common structure of the proofs of (\sqrt{2}), (\sqrt{3}), and (\sqrt{5})?
Correct answer: A
Step 1: First assume the number rational and write it as (\frac{p}{q}) in lowest form. Step 2: Squaring gives a divisibility equation. Step 3: Finally, a common factor in numerator and denominator gives the contradiction.
In the proof of (\sqrt{2}), after assuming (\sqrt{2}=\frac{a}{b}) in lowest form, (a^2=2b^2) is obtained. Which conclusion comes first according to proof order?
Correct answer: A
Step 1: In (a^2=2b^2), the right side has factor (2). Step 2: So first (a^2) is called even, and then (a) is proved even. Step 3: Do not change the order of conclusions in exams.
Assume (\sqrt{3}) is rational and write (\sqrt{3}=\frac{p}{q}). What is the correct reasoning about (p) from (p^2=3q^2)?
Correct answer: B
Step 1: From (p^2=3q^2), (p^2) is divisible by (3). Step 2: Since (3) is prime, (p) is also divisible by (3). Step 3: Use the prime rule to move from square to original number.
If (\sqrt{5}=\frac{m}{n}) is in lowest form and (m^2=5n^2), what is the next aim after writing (m=5k)?
Correct answer: B
Step 1: (m=5k) shows factor (5) in (m). Step 2: Substituting it in (m^2=5n^2) gives (n^2=5k^2). Step 3: Then (n) is also proved divisible by (5), giving contradiction.
If (a) and (b) are coprime, which result will immediately give a contradiction?
Correct answer: B
Step 1: Coprime numbers have no common factor except (1). Step 2: If both are divisible by (3), then (3) is a common factor. Step 3: So it contradicts their being coprime.
In the proof of (\sqrt{2}), why is saying (b) is even immediately after proving (a) even an incomplete reasoning?
Correct answer: A
Step 1: (a) being even does not automatically make (b) even. Step 2: After substituting (a=2k), we get (b^2=2k^2). Step 3: Only then can (b) be proved even.
In the proof of (\sqrt{3}), if (p=3r) and (q=3s), what is definite about (\gcd(p,q))?
Correct answer: A
Step 1: (p=3r) and (q=3s) show factor (3) in both. Step 2: So their greatest common divisor cannot remain (1) and is at least (3). Step 3: This contradicts lowest form.
In the proof of (\sqrt{5}), which conclusion cannot be drawn immediately from (p^2=5q^2)?
Correct answer: D
Step 1: From (p^2=5q^2), first (p^2), then (p), is proved divisible by (5). Step 2: Only after putting (p=5k) do we get (q^2=5k^2). Step 3: So divisibility of (q) is not immediate.
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