When (\sqrt{5}=\frac{p}{q}) is squared, what does the left side become?
Correct answer: B
Step 1: The square of (\sqrt{5}) is (5). Step 2: So the left side becomes ((\sqrt{5})^2=5). Step 3: A square and square root cancel each other.
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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Easy · Level 6
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- (\sqrt{5})
- (5)
- (25)
- (\frac{5}{q})
Easy · Level 6
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- If a prime divides a square, it also divides the original number
- Every prime number is a perfect square
- Every square root is rational
- If a number is positive, it is prime
Easy · Level 6
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- (\sqrt{2}) is rational
- (\sqrt{2}) is irrational
- (\sqrt{2}=2)
- (\sqrt{2}=0)
Easy · Level 6
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- Both (p) and (q) are divisible by (2)
- Both (p) and (q) are divisible by (3)
- Both (p) and (q) are divisible by (5)
- Both (p) and (q) are equal to (9)
Easy · Level 6
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- To show that (q) is also divisible by (5)
- To show that (q=1)
- To show that (p=q)
- To show that (\sqrt{5}=5)
Easy · Level 6
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- Because both are (2)
- Because (\frac{p}{q}) is taken in lowest form
- Because both are zero
- Because both are irrational
Easy · Level 6
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- They will not remain coprime
- They will definitely remain coprime
- Both will become zero
- Both will become irrational
Easy · Level 6
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- Assuming rational makes numerator and denominator of a lowest-form fraction both even
- Because (2) is negative
- Because (\sqrt{2}=2)
- Because every square root is an integer
Easy · Level 6
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- From (p^2=3q^2), (p^2) is divisible by (3)
- Since (p^2) is divisible by (3), (p) is divisible by (3)
- Since (p) is divisible by (3), (p=3k)
- From (p^2=3q^2), (p=3q)
Easy · Level 6
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- Integer
- Irrational number
- Square root
- Necessarily negative
Easy · Level 6
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- Assume rational, square, find both even, contradiction
- Assume both even, then find square root
- Write decimal first, then assume answer
- First write (\sqrt{2}=2)
Easy · Level 6
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- (p=5k)
- (p=2k)
- (p=q)
- (p=k+1)
Easy · Level 6
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- The original statement is true
- The original statement is false
- No conclusion is obtained
- The number is zero
Easy · Level 6
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- In the proof of irrationality of (\sqrt{2})
- In the proof of irrationality of (\sqrt{3})
- In the proof of irrationality of (\sqrt{5})
- In the rationality of (\sqrt{9})
Easy · Level 6
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- If (p^2) is even, then (p) is even
- If (p^2) is even, then (p) is odd
- If (p^2) is even, then (p=1)
- If (p^2) is even, then (q=0)
Easy · Level 6
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- Therefore (\sqrt{3}) is irrational
- Therefore (\sqrt{3}=3)
- Therefore (3) is a perfect square
- Therefore (p=q)
Easy · Level 6
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- They have a common factor other than (1)
- Both are integers
- Both are positive
- Both are written in a fraction
Easy · Level 6
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- Their being coprime
- Their being integers
- Their being positive
- Their being in a fraction
Easy · Level 6
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- In all three, we first assume rationality and write a lowest-form fraction
- In all three, we first assume the number is zero
- In all three, we only write decimal values
- In all three, we take the square root equal to the number under it
Easy · Level 6
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- The assumption led to a contradiction, so the given number is irrational
- The decimal value is enough
- It is not necessary to write numerator and denominator
- Squaring is not necessary
Easy · Level 6
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- (q) is even
- (q) is odd
- (q=1)
- (q) is irrational
Easy · Level 6
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- (q) is divisible by (3)
- (q) is divisible by (2)
- (q=3)
- (q) is zero
Easy · Level 6
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- (q) is divisible by (5)
- (q) is divisible by (2)
- (q=k)
- (q) is not rational
Easy · Level 6
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- Getting (p=2m) and (q=2n)
- (\sqrt{2}) being positive
- (p) and (q) being integers
- (q\neq 0)
Easy · Level 6
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- Assume rational and show the same prime factor in both numerator and denominator
- Find the decimal value and write the answer
- Treat the square root as an integer
- Assume numerator and denominator are equal
Which statement correctly states the prime factor rule?
Correct answer: A
Step 1: Prime factors in a square occur in pairs. Step 2: If a prime divides (p^2), it also divides (p). Step 3: This rule is needed in the proofs of (\sqrt{3}) and (\sqrt{5}).
What is proved at the end in the proof of (\sqrt{2})?
Correct answer: B
Step 1: Assuming rationality makes both (p) and (q) even. Step 2: This contradicts their being coprime. Step 3: Therefore the initial assumption is false and (\sqrt{2}) is irrational.
Which option correctly states the final contradiction in the proof of (\sqrt{3})?
Correct answer: B
Step 1: In the proof of (\sqrt{3}), both (p) and (q) are found divisible by (3). Step 2: But they were assumed coprime at the beginning. Step 3: This is the final contradiction.
In the proof of (\sqrt{5}), after writing (p=5k), what is the next aim?
Correct answer: A
Step 1: First, (p) is found divisible by (5). Step 2: Substituting (p=5k) gives divisibility by (5) for (q) too. Step 3: A common factor in both creates the contradiction.
What is the main reason for writing (p) and (q) coprime in the proof of (\sqrt{2})?
Correct answer: B
Step 1: A rational number is taken as a fraction in lowest form. Step 2: In lowest form, numerator and denominator are coprime. Step 3: Finding a common factor later contradicts this condition.
If both (p) and (q) are divisible by (5), what happens to their being coprime?
Correct answer: A
Step 1: If both are divisible by (5), then (5) is a common factor. Step 2: Coprime numbers should not have a common factor other than (1). Step 3: So this situation goes against being coprime.
Which option is the correct short reason for the irrationality of (\sqrt{2})?
Correct answer: A
Step 1: Assume (\sqrt{2}) rational and write it in lowest form. Step 2: The proof gives both numerator and denominator even. Step 3: This contradicts lowest form, so (\sqrt{2}) is irrational.
Which statement is a wrong step in the proof of (\sqrt{3})?
Correct answer: D
Step 1: From (p^2=3q^2), (p^2) is divisible by (3). Step 2: This gives (p) divisible by (3), but we cannot directly write (p=3q). Step 3: The correct way is to write (p=3k).
In the proof of (\sqrt{2}), after (p) is found even, what type of number is (k) in (p=2k)?
Correct answer: A
Step 1: An even integer is written as (2) times an integer. Step 2: Therefore in (p=2k), (k) is an integer. Step 3: Mentioning the type of (k) makes the proof clear.
Which option gives the correct order of the proof of (\sqrt{2})?
Correct answer: A
Step 1: First assume (\sqrt{2}) is rational. Step 2: After squaring, both (p) and (q) are found even. Step 3: Both being even contradicts the coprime condition.
In the proof of (\sqrt{5}), after (p^2=5q^2), what is the correct form for (p)?
Correct answer: A
Step 1: From (p^2=5q^2), (p^2) is divisible by (5). Step 2: So (p) is also divisible by (5) and is written as (p=5k). Step 3: Choose the correct factor according to the number.
If the rational assumption leads to a contradiction, what is the conclusion about the original statement?
Correct answer: A
Step 1: In contradiction, we work with the opposite assumption. Step 2: If that assumption becomes impossible, the original statement is true. Step 3: So when rationality fails, irrationality is proved.
In which proof are both (p) and (q) found divisible by (2)?
Correct answer: A
Step 1: In the proof of (\sqrt{2}), we get (p^2=2q^2). Step 2: This makes both (p) and (q) divisible by (2), that is even. Step 3: The common factor (2) creates the contradiction.
Which statement is correct for the proof of (\sqrt{2})?
Correct answer: A
Step 1: The square of an even number is even and the square of an odd number is odd. Step 2: So if (p^2) is even, (p) is also even. Step 3: This rule is used in the proof of (\sqrt{2}).
Which option is the correct final sentence to complete the proof of (\sqrt{3})?
Correct answer: A
Step 1: The rational assumption makes both (p) and (q) divisible by (3). Step 2: This contradicts the coprime condition. Step 3: Therefore the final conclusion is that (\sqrt{3}) is irrational.
Which option shows that (p) and (q) are not coprime?
Correct answer: A
Step 1: Coprime numbers have only (1) as a common factor. Step 2: If any common factor other than (1) is found, they are not coprime. Step 3: This contradiction is searched for in irrationality proofs.
In the proof of (\sqrt{5}), both (p) and (q) are found divisible by (5). This is against what?
Correct answer: A
Step 1: At the beginning, (p) and (q) were assumed coprime. Step 2: If both are divisible by (5), then (5) becomes a common factor. Step 3: So this goes against their being coprime.
Which statement is correct for the proofs of (\sqrt{2}), (\sqrt{3}), and (\sqrt{5})?
Correct answer: A
Step 1: All three proofs are based on contradiction. Step 2: At the start, the number is assumed rational and written as a lowest-form fraction. Step 3: Then a common factor gives a contradiction.
In an exam, what should be clearly written in the last line of an irrationality proof?
Correct answer: A
Step 1: The proof starts with the rational assumption. Step 2: At the end, that assumption contradicts the coprime condition. Step 3: In the last line, clearly write both the contradiction and irrationality.
In the proof of (\sqrt{2}), if (q^2=2k^2) is obtained, what conclusion follows about (q)?
Correct answer: A
Step 1: From (q^2=2k^2), (q^2) is divisible by (2). Step 2: If the square of an integer is even, the integer is also even. Step 3: So (q) is even, which helps form the contradiction.
In the proof of (\sqrt{3}), if (q^2=3k^2), what is the correct conclusion about (q)?
Correct answer: A
Step 1: From (q^2=3k^2), (q^2) is divisible by (3). Step 2: Since (3) is prime, (q) is also divisible by (3). Step 3: This shows a common factor in (p) and (q).
In the proof of (\sqrt{5}), if (q^2=5k^2) is obtained, what is the next correct conclusion?
Correct answer: A
Step 1: From (q^2=5k^2), (q^2) is divisible by (5). Step 2: Since (5) is prime, (q) is also divisible by (5). Step 3: Having (5) in both (p) and (q) contradicts the coprime condition.
Which option correctly shows the breaking of the coprime condition in the proof of (\sqrt{2})?
Correct answer: A
Step 1: (p=2m) and (q=2n) mean both are divisible by (2). Step 2: Then (2) becomes their common factor. Step 3: A lowest-form fraction should not have such a common factor.
Which option best describes the common idea in the proofs of (\sqrt{3}) and (\sqrt{5})?
Correct answer: A
Step 1: Factor (3) works in (\sqrt{3}) and factor (5) works in (\sqrt{5}). Step 2: The rational assumption makes that same factor appear in both numerator and denominator. Step 3: This common factor contradicts lowest form.
QUIZ COMPLETE