Question 180067
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I can explain these things in my words, but if you are going to use this for your homework, you'll have to change them and still manage to communicate the same ideas in order for you to properly answer the questions.


*[tex \Large (a + b)^2 \neq a^2 + b^2]


Obviously, these two expressions <i><b>are</b></i> equal if either <i><b>a</b></i>, <i><b>b</b></i>, or both are 0.  The question is, do there exist any non-zero values for <i><b>a</b></i> and <i><b>b</b></i> that exhibit the property *[tex \Large (a + b)^2 = a^2 + b^2]? 


Let's assume that there exists a number *[tex \Large a \neq 0] and a number *[tex \Large b \neq 0] such that *[tex \Large (a + b)^2 = a^2 + b^2].


We can add *[tex \Large -b^2] to both sides:


*[tex \Large \text{          }(a + b)^2 - b^2 = a^2].


Then we can factor the difference of two squares on the left:


*[tex \Large \text{          } ((a + b) - b)((a + b) + b) = a^2].


Collect terms and distribute:


*[tex \Large \text{          } a(a + 2b) = a^2].


*[tex \Large \text{          } a^2 + 2ab = a^2].


Now add *[tex \Large -a^2] to both sides:


*[tex \Large \text{          } 2ab = 0].


According to the Zero Product Rule, if *[tex \Large 2ab = 0] then either *[tex \Large a = 0] or *[tex \Large b = 0], but that contradicts our original assumption that *[tex \Large a \neq 0] and *[tex \Large b \neq 0].  Hence, <i>reductio ad absurdum</i> there are no non-zero values for <i>a</i> and <i>b</i> such that *[tex \Large (a + b)^2 = a^2 + b^2].  Therefore, *[tex \Large (a + b)^2 \neq a^2 + b^2 \text{       } \math \forall a \neq 0 \text{ and } \math b \neq 0] 


Having said all of that, finding a numerical example to illustrate *[tex \Large (a + b)^2 \neq a^2 + b^2] should be a fairly simple task.  I'll leave that part to you.


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Add -5 to the inequality 7 > 1


*[tex \Large -5 + 7 > -5 + 1]


*[tex \Large 2 > -4] which is a true statement without changing the sense of the inequality.


Then divide both sides of the inequality -25 > -30 by -6.


*[tex \Large \frac {-25}{-6} > \frac {-30}{-6}]



*[tex \Large \frac {25}{6} > \frac {30}{6}]


Which is now a false statement.  Since we divided by a number less than zero, the sense of the inequality must be reversed.  This is also true when you multiply by a negative, since dividing by a number is the same as multiplying by the reciprocal.



*[tex \Large \frac {25}{6} < \frac {30}{6}]


Having reversed the sense of the inequality, the statement is again true.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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