Math
posted by Wenny .
What is wrong with the following proof. You must explain your answer in words.
a>3
3a>3(3)
3aa^2>9a^2
a(3a)>(3a)(3+a)
a>3+a
0>3
the problem with the proof is that the fourth step needs the sign to be reverse...but why?
i looked at step 3 and there is nothing negative to reverse the sigh for...

From the first statement, a must be a positive number. In fact, it must exceed 3. (This is very important later). The second and third statements are correct, since you are either multiplying by positive numbers or just factoring. The fourth statement is incorrect because 3a is a negative number.The directon of the > sign must change.
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