a+b=c+d and a<c Use proof by contradiction to show that b>d.
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Ok, let's take values that satisfices condition that a+b = c+d and a<c
So, we may take, a=3, b =2, c= 4 and d =1
So, it satisfies as 3 + 2 = 5
And also 4 + 1 = 5.
Again, 4 > 3, so a<c.
Clearly, 2>1, thus b>d.
If helps, please mark it as brainliest ^_^
So, we may take, a=3, b =2, c= 4 and d =1
So, it satisfies as 3 + 2 = 5
And also 4 + 1 = 5.
Again, 4 > 3, so a<c.
Clearly, 2>1, thus b>d.
If helps, please mark it as brainliest ^_^
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