-
PROOFS IN MATH
Suppose a + b = c + d, and a < c. Use proof by contradiction to show b > d.
-
2 Replies
-
Assume that b < d i.e b-d <0 ….(1)
Given that a <c i.e a-c < 0and a+b=c+d
The given equation can be written as
a-c=d-b
We know that a-c <0
Therefore d-b <0 …..(2)
Equation (2) contradicts our assumption
Which implies that our assumption has gone wrong
Therefore d <b
-
To prove that b > d given the conditions a + b = c + d and a < c, we will use proof by contradiction.
Assume, for the sake of contradiction, that b ≤ d.
Since a < c, we can subtract a from both sides of the equation a + b = c + d to get b = c – a + d.
Now, let’s consider the case where b = d.
If b = d, we can substitute d for b in the equation b = c – a + d, which gives d = c – a + d.
By subtracting d from both sides of the equation, we have 0 = c – a, or a = c.
This contradicts the given condition that a < c. Therefore, our assumption that b ≤ d must be false.
Hence, we can conclude that b > d.
In simpler terms, if we assume that b is not greater than d (i.e., b ≤ d) and follow the logical steps, we reach a contradiction where a would be equal to c. This contradicts the given condition a < c, proving that our assumption is incorrect. Therefore, b must be greater than d.
In summary, using proof by contradiction, we have demonstrated that if a + b = c + d and a < c, then it follows that b > d.
Log in to reply.