Which statement is directly correct from (p^2=5q^2)?
Correct answer: A
Step 1: In (p^2=5q^2), (p^2) equals (5q^2). Step 2: So (p^2) definitely has factor (5). Step 3: Divisibility of (q^2) comes in a later step.
Mathematics
Proof of irrationality of √2, √3, √5
√2, √3 और √5 की अपरिमेयता का प्रमाण
In this Class 10 Mathematics topic from the Real Numbers chapter, students learn how to prove that √2, √3 and √5 are irrational numbers. The proof begins by assuming that a square root can be written as a fraction p/q in lowest terms, then uses prime divisibility and the resulting contradiction to reject that assumption. Students also strengthen their understanding of rational and irrational numbers, prime factorisation, and the logical structure used in mathematical proofs.
TOPIC PRACTICE
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Medium · Level 2
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- (p^2) is divisible by (5)
- (q^2) is divisible by (5)
- (p=q)
- (q=5p)
Medium · Level 2
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- The only common factor of (p) and (q) is (1)
- Both (p) and (q) are even
- (p=q)
- Both (p) and (q) are (2)
Medium · Level 2
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- Squaring both sides
- Adding (3) to both sides
- Subtracting (q) from both sides
- Multiplying both sides by zero
Medium · Level 2
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- (\frac{p}{q}) was not in lowest form
- (\frac{p}{q}) is necessarily an integer
- (\sqrt{5}=5)
- (q=0)
Medium · Level 2
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- In (\sqrt{2}), common factor (2) is found; in (\sqrt{5}), common factor (5) is found
- No common factor is found in either
- In (\sqrt{2}), (5) is found; in (\sqrt{5}), (2) is found
- In both, (q=0) is found
Medium · Level 2
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- Both use the prime factor divisibility rule
- Both use only the even number rule
- In both, denominator is assumed zero
- In both, the square root is assumed integer
Medium · Level 2
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- Because the square of an odd number is odd
- Because every number is even
- Because (p=q)
- Because (q=0)
Medium · Level 2
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- (\sqrt{5}) is irrational
- (\sqrt{5}) is rational
- (\sqrt{5}=5)
- (5) is a perfect square
Medium · Level 2
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- (b) is divisible by (3)
- (b) is divisible by (2)
- (b=1)
- (b) is zero
Medium · Level 2
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- It is not in lowest form
- It is necessarily (1)
- It is always an integer
- It is zero
Medium · Level 2
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- The denominator in (\frac{p}{q}) cannot be zero
- (q) must always be (2)
- (q) must always be even
- (q) must be irrational
Medium · Level 2
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- (\sqrt{2})
- (\sqrt{3})
- (\sqrt{5})
- (\sqrt{9})
Medium · Level 2
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- (p) and (q) are not coprime
- (p) and (q) are not integers
- (\sqrt{2}) is negative
- (q=0)
Medium · Level 2
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- To show (q) is also divisible by (5)
- To show (p=q)
- To show (\sqrt{5}=5)
- To show (q=0)
Medium · Level 2
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- Stopping after only writing (p^2=3q^2)
- Showing both (p) and (q) divisible by (3)
- Writing contradiction with coprime condition
- Finally writing (\sqrt{3}) is irrational
Medium · Level 2
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- (\sqrt{2}=\frac{p}{q}), where (p), (q) are integers and (q\neq 0)
- (\sqrt{2}=p+q), where (p), (q) are integers
- (\sqrt{2}=pq), where (p), (q) are positive
- (\sqrt{2}=0)
Medium · Level 2
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- A clear contradiction is formed when a common factor is found
- Decimal value is obtained immediately
- Denominator becomes zero
- (p=q) is proved
Medium · Level 2
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- (p^2=5q^2), (p=5k), (25k^2=5q^2), (q^2=5k^2)
- (p^2=5q^2), (p=5q), (q=5)
- (p^2=5q^2), (q=0)
- (p^2=5q^2), (\sqrt{5}=5)
Medium · Level 2
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- (p^2=4k^2)
- (4k^2=2q^2)
- (q^2=2k^2)
- (p^2=2k^2)
Medium · Level 2
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- If a prime divides a square, it divides the original number
- If a number is positive, it is a perfect square
- Every fraction is not rational
- Every square root is an integer
Medium · Level 2
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- When (p) and (q) in lowest form are both proved even
- When (p) and (q) are integers
- When (q\neq 0)
- When (\sqrt{2}) is positive
Medium · Level 2
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- The square root of a rational number is not always rational
- Every rational number is zero
- (5) is not rational
- (\sqrt{5}=5)
Medium · Level 2
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- (\sqrt{3}) is irrational because assuming rational makes both (p) and (q) divisible by (3)
- (\sqrt{3}) is rational because (3) is an integer
- (\sqrt{3}=3) because the radical disappears
- (\sqrt{3}) is an integer because (3) is prime
Medium · Level 2
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- Taking the square root equal to the number under it
- Assuming rational and writing (\frac{p}{q})
- Squaring both sides
- Showing contradiction using common factor
Medium · Level 2
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- This contradicts our rational assumption, hence the given number is irrational
- The approximate decimal value is enough
- Hence the given number is a perfect square
- Hence the denominator is zero
Which option gives the correct meaning of the coprime condition for (p) and (q) in the proof of (\sqrt{2})?
Correct answer: A
Step 1: Coprime means two numbers have no common factor except (1). Step 2: So finding both even breaks this meaning. Step 3: Understanding the definition makes the proof easier.
In the proof of (\sqrt{3}), which step is done to remove the square root?
Correct answer: A
Step 1: (\sqrt{3}) contains a square root. Step 2: To remove it, we square both sides and get (3=\frac{p^2}{q^2}). Step 3: Choose the correct algebraic operation to remove the radical.
If in the proof of (\sqrt{5}), both (p) and (q) are proved divisible by (5), what does it mean?
Correct answer: A
Step 1: If both are divisible by (5), the fraction has common factor (5). Step 2: Such a fraction can be reduced further. Step 3: This contradicts the assumption of lowest form.
Which option correctly shows a difference between the proofs of (\sqrt{2}) and (\sqrt{5})?
Correct answer: A
Step 1: In the proof of (\sqrt{2}), (2) is the key factor. Step 2: In the proof of (\sqrt{5}), (5) is the key factor. Step 3: Pay attention to the number under the root in each proof.
Which option shows a similarity between the proofs of (\sqrt{3}) and (\sqrt{5})?
Correct answer: A
Step 1: In (\sqrt{3}), (3) is prime; in (\sqrt{5}), (5) is prime. Step 2: Both proofs use divisibility from square to original number. Step 3: This is their main common logic.
A student writes in the proof of (\sqrt{2}) that (p^2) is even so (p) is even. Why is this step correct?
Correct answer: A
Step 1: If (p) were odd, then (p^2) would also be odd. Step 2: But (p^2) is even, so (p) cannot be odd. Step 3: Thus (p) must be even.
If assuming (\sqrt{5}) rational finally gives a contradiction, which conclusion is correct?
Correct answer: A
Step 1: In contradiction, the opposite assumption is taken. Step 2: If the rational assumption becomes impossible, it is false. Step 3: Therefore (\sqrt{5}) is proved irrational.
Which statement correctly tells the final conclusion about (b) in the proof of (\sqrt{3})?
Correct answer: A
Step 1: After substituting (a=3k), we get (b^2=3k^2). Step 2: Hence (b^2) is divisible by (3). Step 3: By the prime rule, (b) is also divisible by (3).
In the proof of (\sqrt{2}), if both (p) and (q) are even, what can be said about the fraction (\frac{p}{q})?
Correct answer: A
Step 1: Both even means numerator and denominator have common factor (2). Step 2: Such a fraction can be reduced by (2). Step 3: So it cannot be in lowest form.
Which statement correctly explains why (q\neq 0) is needed in the proof of (\sqrt{2})?
Correct answer: A
Step 1: A rational number is written as (\frac{p}{q}). Step 2: The denominator of a fraction cannot be zero. Step 3: Therefore (q\neq 0) must be written.
In a proof, (p^2=3q^2) is obtained. This is related to the irrationality proof of which square root?
Correct answer: B
Step 1: Assuming (\sqrt{n}=\frac{p}{q}) and squaring gives (p^2=nq^2). Step 2: Here (n=3), so it relates to (\sqrt{3}). Step 3: Identify the square root from the factor in the equation.
In a proof, from (p^2=2q^2), we get (p=2r) and then (q=2s). What contradiction does this give?
Correct answer: A
Step 1: (p=2r) and (q=2s) mean both are divisible by (2). Step 2: So they cannot be coprime. Step 3: But they were assumed coprime at the start, which is the contradiction.
In the proof of (\sqrt{5}), (p) is found divisible by (5) from (p^2=5q^2). What is the purpose of putting (p=5k)?
Correct answer: A
Step 1: First (p) is found to have factor (5). Step 2: Substituting (p=5k) in the equation gives (q^2=5k^2). Step 3: This proves (q) is also divisible by (5).
Which statement leaves the proof of (\sqrt{3}) incomplete?
Correct answer: A
Step 1: (p^2=3q^2) is a middle step, not the end. Step 2: After this, both (p) and (q) must be shown divisible by (3). Step 3: The proof is incomplete without contradiction and conclusion.
Which option is the correct use of the definition of rational number in the proof of (\sqrt{2})?
Correct answer: A
Step 1: A rational number is written as a ratio of two integers. Step 2: The denominator cannot be zero, so (q\neq 0) is necessary. Step 3: In lowest form, (p) and (q) are also taken coprime.
If (\sqrt{3}) is assumed rational and written as (\frac{p}{q}), what is the benefit of taking (\frac{p}{q}) in lowest form?
Correct answer: A
Step 1: In lowest form, (p) and (q) are coprime. Step 2: When the proof shows both divisible by (3), this becomes impossible. Step 3: Thus lowest form helps show contradiction.
Which option shows the correct path to prove (q) divisible by (5) in the proof of (\sqrt{5})?
Correct answer: A
Step 1: From (p^2=5q^2), (p) is divisible by (5), so (p=5k). Step 2: Substitution gives (25k^2=5q^2), then (q^2=5k^2). Step 3: This proves (q) is also divisible by (5).
Which statement is a wrong simplification after putting (p=2k) in the proof of (\sqrt{2})?
Correct answer: D
Step 1: If (p=2k), then (p^2=(2k)^2). Step 2: Its correct value is (4k^2), not (2k^2). Step 3: Square the coefficient while squaring.
In the proof of (\sqrt{3}), from (p^2=3q^2), (p) is divisible by (3). This is based on which rule?
Correct answer: A
Step 1: From (p^2=3q^2), (p^2) is divisible by (3). Step 2: (3) is prime, so the prime divisibility rule applies. Step 3: Therefore (p) is also divisible by (3).
In which situation is the rational assumption for (\sqrt{2}) proved false?
Correct answer: A
Step 1: In lowest form, (p) and (q) should be coprime. Step 2: If both are proved even, both have common factor (2). Step 3: This is impossible, so the rational assumption is false.
A student wrote that (\sqrt{5}) is rational because (5) is rational. What is the mistake in this reasoning?
Correct answer: A
Step 1: (5) is rational, but it is not a perfect square. Step 2: The square root of a non-perfect square need not be rational, and (\sqrt{5}) is irrational. Step 3: Check a number and its square root separately.
Which option gives both the correct final conclusion and reason in the proof of (\sqrt{3})?
Correct answer: A
Step 1: Assuming (\sqrt{3}) rational gives (p^2=3q^2). Step 2: This proves both (p) and (q) divisible by (3). Step 3: This contradicts coprime condition, so (\sqrt{3}) is irrational.
Which option is a wrong method in all three proofs?
Correct answer: A
Step 1: Treating (\sqrt{2}), (\sqrt{3}), and (\sqrt{5}) as (2), (3), and (5) is wrong. Step 2: The correct method assumes rationality, writes a fraction, and squares. Step 3: Do not write a square root equal to the number under it.
In an exam, what is the most important final line while proving the irrationality of (\sqrt{2}), (\sqrt{3}), or (\sqrt{5})?
Correct answer: A
Step 1: The proof starts with the rational assumption. Step 2: At the end, a contradiction appears with the coprime condition. Step 3: The final line should clearly state the contradiction and irrationality conclusion.
QUIZ COMPLETE