Lesson Worksheet: Proof by Contradiction Mathematics

In this worksheet, we will practice describing what proof by contradiction is and using it to determine whether a conjecture is true or false.

Q1:

A student has tried to prove the following statement by contradiction: β€œFor all even integers π‘š and 𝑛, their product π‘šπ‘› is divisible by 4.” The student’s working is shown below.

  1. Assumption: All integers 𝑛 such that π‘›οŠ¨ is odd are even.
  2. 𝑛 is even; so, write 𝑛=2π‘˜
  3. Then, 𝑛=(2π‘˜)=4π‘˜=2ο€Ή2π‘˜ο…οŠ¨οŠ¨οŠ¨οŠ¨, so π‘›οŠ¨ is even.
  4. This contradicts the assumption that π‘›οŠ¨ is odd.
  5. Therefore, if π‘›οŠ¨ is odd, then 𝑛 must be odd.

Identify the error in their working.

  • AIn the second sentence, they should have written π‘š=2π‘˜+1 and 𝑛=2𝑙+1.
  • BIn the final sentence, they have drawn the wrong conclusion.
  • CIn the fourth sentence, they have stated the wrong assumption.
  • DIn the third sentence, the algebra is incorrect.
  • EIn the first sentence, they have given the incorrect negation of the statement they want to prove.

Q2:

Choose the statement that is the negation of β€œall multiples of 5 are odd.”

  • AAll odd numbers are multiples of 5.
  • BNo multiple of 5 is odd.
  • CAll multiples of 5 are even.
  • DAt least one multiple of 5 is even.
  • EMore than one multiple of 5 is even.

Q3:

What is the negation of each of the following statements?

All swans are white.

  • AAt least one swan is not white.
  • BMore than one swan is not white.
  • CNo swans are white.
  • DAll swans are not white.
  • EAll swans are black.

If π‘Ž and 𝑏 are prime numbers, then (π‘Žπ‘+1) is a prime number.

  • AThere exist prime numbers π‘Ž and 𝑏 such that (π‘Žπ‘+1) is not prime.
  • BIf (π‘Žπ‘+1) is a prime number, then π‘Ž and 𝑏 are prime numbers.
  • CIf π‘Ž and 𝑏 are not prime numbers, then there exists a number of the form (π‘Žπ‘+1) that is prime.
  • DIf π‘Ž and 𝑏 are not prime numbers, then (π‘Žπ‘+1) is not prime.
  • EIf π‘Ž and 𝑏 are prime numbers, then (π‘Žπ‘+1) is not prime.

All odd numbers are either multiples of 3 or can be written in the form 2βˆ’1.

  • AThere is an odd number that is neither a multiple of 3 nor can be written in the form 2βˆ’1.
  • BAll numbers that are either multiples of 3 or can be written in the form 2βˆ’1 are odd.
  • CNo numbers that are either multiples of 3 or can be written in the form 2βˆ’1 are odd.
  • DAll even numbers are either multiples of 3 or can be written in the form 2βˆ’1.
  • ENo odd numbers are either multiples of 3 or can be written in the form 2βˆ’1.

Q4:

Madison wants to prove by contradiction that if 𝑝+π‘ž is an irrational number, then at least one of 𝑝 and π‘ž is irrational.

Which of the following assumptions should Madison use at the start of her proof?

  • AAssume that 𝑝+π‘ž is an irrational number and that neither 𝑝 nor π‘ž is irrational.
  • BAssume that 𝑝+π‘ž is an irrational number and that both 𝑝 and π‘ž are irrational.
  • CAssume that 𝑝+π‘ž is a rational number and that neither 𝑝 nor π‘ž is irrational.
  • DAssume that 𝑝+π‘ž is a rational number and that at least one of 𝑝 and π‘ž is irrational.
  • EAssume that at least one of 𝑝 and π‘ž is an irrational number and that 𝑝+π‘ž is irrational.

Which of the following statements is the contradiction that Madison should obtain near the end of her proof?

  • AAssuming that neither 𝑝 nor π‘ž is irrational leads to the contradiction that 𝑝+π‘ž is rational.
  • BAssuming that neither 𝑝 nor π‘ž is irrational leads to the contradiction that 𝑝+π‘ž is irrational.
  • CAssuming that 𝑝 or π‘ž is irrational leads to the contradiction that 𝑝+π‘ž is irrational.
  • DAssuming that both 𝑝 and π‘ž are irrational leads to the contradiction that 𝑝+π‘ž is rational.
  • EAssuming that 𝑝+π‘ž is irrational leads to the contradiction that neither 𝑝 nor π‘ž is irrational.

Madison’s classmate makes this statement: β€œIf 𝑝+π‘ž is a rational number, then at least one of 𝑝 and π‘ž must be rational.”

Disprove the statement by choosing a suitable counterexample from the following.

  • A𝑝=√2 and π‘ž=πœ‹βˆ’βˆš2
  • B𝑝=√2 and √3
  • C𝑝=2 and π‘ž=βˆ’3
  • D𝑝=√2 and π‘ž=2
  • E𝑝=√2 and π‘ž=2βˆ’βˆš2

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