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Hard · Level 17 · common contradiction,irrationality proof,hard,class 10View options
Finding a common factor in numerator and denominator of a lowest-form fraction
The square root being positive
Denominator being non-zero
Numerator and denominator being integers
Hard · Level 17 · sqrt3 proof,q divisibility,hard,class 10View options
(q^2) is divisible by (3) and (3) is prime
(q=3) already
(k=q)
(q) is zero
Hard · Level 17 · sqrt5 proof,fraction reduction,hard,class 10View options
(\frac{p}{q}=\frac{5k}{5r}=\frac{k}{r})
(\frac{p}{q}=\frac{k}{5r})
(\frac{p}{q}=5kr)
(\frac{p}{q}=\frac{25k}{r})
Hard · Level 17 · sqrt2 proof,final contradiction,hard,class 10View options
(a) and (b) were assumed coprime, but both turned out even
(\sqrt{2}) is positive, so there is contradiction
(b\neq 0), so there is contradiction
(a) and (b) are integers, so there is contradiction
Hard · Level 17 · sqrt3 proof,squaring error,hard,class 10View options
Squaring both sides gives (3=\frac{p^2}{q^2})
Squaring both sides gives (3=\frac{p}{q^2})
(3=\frac{p}{q}) is correct without squaring
Squaring both sides gives (9=\frac{p}{q})
Hard · Level 17 · sqrt5 proof,prime condition,hard,class 10View options
(5) is prime
(5) is even
(q=0)
(\sqrt{5}=5)
Hard · Level 17 · sqrt2 proof,gcd,hard,class 10View options
(\gcd(a,b)) is at least (2)
(\gcd(a,b)) will remain (1)
(\gcd(a,b)=0)
(\gcd(a,b)=a+b)
Hard · Level 17 · prime divisibility,irrationality proof,hard,class 10View options
To conclude (r\mid x)
To conclude (x\mid r)
To conclude (x=r^2)
To conclude (x=0)
Hard · Level 17 · sqrt5 proof,incomplete proof,hard,class 10View options
Stopping after only writing (p^2=5q^2)
Showing both (p) and (q) divisible by (5)
Writing contradiction with coprime condition
Finally writing (\sqrt{5}) is irrational
Hard · Level 17 · sqrt2 proof,incomplete conclusion,hard,class 10View options
(a) is even
(a=b)
(b=0)
(\sqrt{2}=2)
Hard · Level 17 · sqrt3 proof,logical order,hard,class 10View options
From (p^2=3q^2), first show (p), then (q), divisible by (3)
Directly writing (q) is divisible by (3) from (p^2=3q^2)
Getting (q^2=3k^2) after putting (p=3k)
Writing contradiction when both (p) and (q) are divisible by (3)
Hard · Level 17 · sqrt5 proof,coprime condition,hard,class 10View options
(p) and (q) are coprime
(q\neq 0)
(p) and (q) are integers
(\sqrt{5}) is positive
Hard · Level 17 · sqrt2 proof,fraction reduction,hard,class 10View options
(\frac{a}{b}) can be reduced to (\frac{m}{n})
(\frac{a}{b}=2)
(\frac{a}{b}=0)
(\frac{a}{b}) is undefined
Hard · Level 17 · sqrt3 proof,algebra mistake,hard,class 10View options
From (p=3k), (p^2=9k^2)
From (9k^2=3q^2), (q^2=3k^2)
From (p=3k), (p^2=3k^2)
From (q^2=3k^2), (q) is divisible by (3)
Hard · Level 17 · sqrt5 proof,final reason,hard,class 10View options
Assuming rational makes both numerator and denominator of the lowest-form fraction divisible by (5)
Because (5) is positive
Because (\sqrt{5}=5)
Because the denominator becomes zero
Hard · Level 17 · squaring role,proof structure,hard,class 10View options
To remove the square root and create a divisibility equation
To make the denominator zero
To prove the number is a perfect square
To find decimal expansion
Hard · Level 17 · sqrt3 proof,lowest form,hard,class 10View options
It cannot remain in lowest form
It must become (3)
It will become zero
It will be undefined
Hard · Level 17 · sqrt2 proof,even square,logic,hardView options
Because if (a) were odd, then (a^2) would also be odd
Because every number is even
Because (a=b)
Because (b=0)
Hard · Level 17 · proof comparison,sqrt2 sqrt5,hard,class 10View options
(\sqrt{2}) gives common factor (2), while (\sqrt{5}) gives common factor (5)
Both give common factor (3)
(\sqrt{2}) gives (5), while (\sqrt{5}) gives (2)
No common factor is found in either
Hard · Level 17 · sqrt3 sqrt5,prime factor,hard,class 10View options
If a prime factor divides a square, it also divides the original number
The denominator must be assumed zero
The square root must be assumed equal to the number inside
Decimal expansion itself is the proof
Question 1HardLevel 17
What is the root cause of contradiction in the proofs of (\sqrt{2}), (\sqrt{3}), and (\sqrt{5})?
Correct answer: A
Step 1: After assuming rationality, the number is written in lowest-form fraction. Step 2: The proof finds the same factor in both numerator and denominator. Step 3: This cannot happen in lowest form, so contradiction occurs.
In the proof of (\sqrt{5}), if (p=5k) and (q=5r), how can (\frac{p}{q}) be reduced?
Correct answer: A
Step 1: If (p=5k) and (q=5r), both numerator and denominator share (5). Step 2: (\frac{5k}{5r}) can be reduced to (\frac{k}{r}). Step 3: This shows the fraction was not in lowest form.
Which option states the final contradiction in the proof of (\sqrt{2}) most correctly?
Correct answer: A
Step 1: In lowest form, (a) and (b) were assumed coprime. Step 2: The proof shows both are even, so both have common factor (2). Step 3: This is the correct final contradiction.
If a student writes (3=\frac{p}{q}) directly from (\sqrt{3}=\frac{p}{q}), what is the correct correction?
Correct answer: A
Step 1: To get (3) from (\sqrt{3}), both sides must be squared. Step 2: The square of a fraction is (\frac{p^2}{q^2}). Step 3: So the correct form is (3=\frac{p^2}{q^2}).
In the proof of (\sqrt{2}), if (a=2k) and (b=2r), which statement about (\gcd(a,b)) is correct?
Correct answer: A
Step 1: (a=2k) and (b=2r) show both are divisible by (2). Step 2: So their greatest common divisor cannot remain (1). Step 3: This breaks the initial coprime condition.
If (r) is prime and (r\mid x^2), how is it used in the proofs of (\sqrt{3}) and (\sqrt{5})?
Correct answer: A
Step 1: If a prime divides a square, it also divides the original number. Step 2: In (\sqrt{3}), this is used for (3); in (\sqrt{5}), it is used for (5). Step 3: This gives a common factor in numerator and denominator.
Which statement leaves the proof of (\sqrt{5}) incomplete?
Correct answer: A
Step 1: (p^2=5q^2) is only a middle step. Step 2: After this, both (p) and (q) must be shown divisible by (5). Step 3: Without contradiction and final conclusion, the proof is incomplete.
In the proof of (\sqrt{2}), which option is a correct but incomplete conclusion?
Correct answer: A
Step 1: From (a^2=2b^2), (a) is proved even. Step 2: But to complete the proof, (b) must also be proved even. Step 3: Only when both are even does contradiction arise with the coprime condition.
Which option disturbs the logical order in the proof of (\sqrt{3})?
Correct answer: B
Step 1: From (p^2=3q^2), first (p) is concluded divisible by (3). Step 2: After substituting (p=3k), (q^2=3k^2) is obtained. Step 3: Therefore jumping directly to (q) is an order mistake.
In the proof of (\sqrt{5}), if (p=5k) and (q=5r) are obtained, which initial condition breaks?
Correct answer: A
Step 1: (p=5k) and (q=5r) show factor (5) in both (p) and (q). Step 2: So they cannot be coprime. Step 3: This breaks the initial lowest-form condition.
In the proof of (\sqrt{2}), (\frac{a}{b}) is in lowest form. If (a=2m) and (b=2n), which conclusion is most suitable?
Correct answer: A
Step 1: (a=2m) and (b=2n) show common factor (2) in numerator and denominator. Step 2: So (\frac{2m}{2n}=\frac{m}{n}). Step 3: This contradicts lowest form.
Which final reason is most accurate in proving (\sqrt{5}) irrational?
Correct answer: A
Step 1: Assume (\sqrt{5}) rational and write it in lowest-form fraction. Step 2: The proof shows both numerator and denominator divisible by (5). Step 3: This contradicts the coprime condition, so (\sqrt{5}) is irrational.
Which statement tells the real role of squaring in the proofs of (\sqrt{2}), (\sqrt{3}), and (\sqrt{5})?
Correct answer: A
Step 1: Squaring (\sqrt{n}=\frac{p}{q}) removes (\sqrt{n}). Step 2: This creates an equation like (p^2=nq^2). Step 3: Divisibility and contradiction start from this equation.
If in the proof of (\sqrt{3}), both (p) and (q) are divisible by (3), what is the effect on the lowest form of (\frac{p}{q})?
Correct answer: A
Step 1: If both are divisible by (3), the fraction has common factor (3). Step 2: Such a fraction can be reduced by (3). Step 3: Therefore the lowest-form assumption breaks.
Which option correctly states the difference between the proofs of (\sqrt{2}) and (\sqrt{5})?
Correct answer: A
Step 1: In (\sqrt{2}), (a^2=2b^2) makes (2) the key factor. Step 2: In (\sqrt{5}), (p^2=5q^2) makes (5) the key factor. Step 3: The number inside the root decides the proof factor.
Which deeper idea is common in the proofs of (\sqrt{3}) and (\sqrt{5})?
Correct answer: A
Step 1: Both (3) and (5) are prime. Step 2: When these factors appear in (p^2), they also appear in (p). Step 3: This idea finally gives a common factor in numerator and denominator.
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