Question 6 of 26intermediate🔧 ApplyLong Answer3 marks

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

Correct Answer

We want to prove that b>db > d. Using proof by contradiction, we start by assuming the negation of this statement, that is, b≤db \leq d.

Given that a+b=c+da + b = c + d, we can rearrange this equation to get:

a−c=d−ba - c = d - b

Since it is given that a<ca < c, we have:

a−c<0a - c < 0

This means:

d−b<0d - b < 0

which implies b>db > d.

However, this result b>db > d contradicts our assumption that b≤db \leq d. Since we have arrived at a contradiction, our assumption must be false. Therefore, b>db > d is true.

Exercise: EXERCISE A1.6 | Q: 1 | (Chapter: 21)
For More Understanding

Explanation

This question tests the understanding of proof by contradiction as explained in Section A1.7 of the textbook. The method requires assuming the negation of what we want to prove and showing this leads to a contradiction.

The textbook states that 'a contradiction occurs when we get a statement pp such that pp is true and ∼p\sim p, its negation, is also true.' Here, assuming b≤db \leq d leads to the contradictory conclusion b>db > d, proving the original statement.

Solution Steps

  1. Step 1: Assume the negation of 'b>db > d', which is 'b≤db \leq d'.

  2. Step 2: From a+b=c+da + b = c + d, rearrange to get a−c=d−ba - c = d - b.

  3. Step 3: Since a<ca < c, we have a−c<0a - c < 0, therefore d−b<0d - b < 0, which means b>db > d.

  4. Step 4: This contradicts our assumption b≤db \leq d, hence b>db > d is proved.

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