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Activity Discussion Math PROOFS IN MATH

• # PROOFS IN MATH

Posted by on June 23, 2023 at 11:25 pm

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

replied 1 year, 1 month ago 3 Members · 2 Replies
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• ### Anagha

Member
June 24, 2023 at 12:22 pm
0

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)

Which implies that our assumption has gone wrong

Therefore d <b

• ### Suryansh

Member
June 24, 2023 at 1:20 pm
0

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 â‰¤ 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 â‰¤ 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 â‰¤ 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.

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