Asked by Jenny
Elizabeth claims that she has proven that 7=5with the following:
7(a+b)-5(a+b)=7c-5c
7a+7b-5a-5b=7c-5c
7a+7b-7c=5a+5b-5c
7(a+b-c)/(a+b-c) = 5(a+b-c)/(a+b-c)
7=5
1.) Explain why this is an invalid proof.
7(a+b)-5(a+b)=7c-5c
7a+7b-5a-5b=7c-5c
7a+7b-7c=5a+5b-5c
7(a+b-c)/(a+b-c) = 5(a+b-c)/(a+b-c)
7=5
1.) Explain why this is an invalid proof.
Answers
Answered by
Reiny
First of all, where does the first line come from?
if it is true, then
7(a+b)-5(a+b)=7c-5c
2(a+b) = 2c
a+b = c
a+b-c = 0
so in your second last line you divided by zero, which is a mathematical No No.
any subsequent conclusion you draw would be bogus.
if it is true, then
7(a+b)-5(a+b)=7c-5c
2(a+b) = 2c
a+b = c
a+b-c = 0
so in your second last line you divided by zero, which is a mathematical No No.
any subsequent conclusion you draw would be bogus.
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