Appendix 1: Proofs in Mathematics | A1.6

Question 1

Suppose a+b=c+da + b = c + d, and a<ca < c. Use proof by contradiction to show b>db > d.

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Solution
Understand the Question
  • Given: a+b=c+da + b = c + d and a<ca < c.
  • Goal: Show that b>db > d using proof by contradiction.
  • Method: In a proof by contradiction, we assume the opposite (negation) of the desired result (bdb \le d) and show that it leads to a logical impossibility or contradicts a given condition.

Step 1 · Assume the Negation of the Statement

To prove b>db > d by contradiction, assume the opposite is true: bdb \le d

Step 2 · Combine Inequalities to Reach a Contradiction

We are given: a<ca < c

Adding the assumption bdb \le d to a<ca < c: a+b<c+da + b < c + d

This contradicts the given statement that: a+b=c+da + b = c + d

Since our assumption leads to a contradiction, the assumption bdb \le d must be false.

Therefore: b>db > d

Answer

Hence, by proof by contradiction, b>db > d.

Common Mistakes
  • Incomplete Negation: Incorrectly assuming the opposite of b>db > d is only b<db < d, forgetting the equality case b=db = d (the complete negation is bdb \le d).
  • Missing Contradiction Link: Not explicitly comparing the derived inequality (a+b<c+da + b < c + d) with the given equation (a+b=c+da + b = c + d) to state where the contradiction occurs.

More questions in A1.6

Q1

Suppose a+b=c+da + b = c + d, and a<ca < c. Use proof by contradiction to show b>db > d.

Q2

Let rr be a rational number and xx be an irrational number. Use proof by contradiction to show that r+xr + x is an irrational number.

Q3

Use proof by contradiction to prove that if for an integer aa, a2a^2 is even, then so is aa.

[Hint : Assume aa is not even, that is, it is of the form 2n+12n + 1, for some integer nn, and then proceed.]

Q4

Use proof by contradiction to prove that if for an integer aa, a2a^2 is divisible by 3, then aa is divisible by 3.

Q5

Use proof by contradiction to show that there is no value of nn for which 6n6^n ends with the digit zero.

Q6

Prove by contradiction that two distinct lines in a plane cannot intersect in more than one point.

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