Question

Does (x+y)^n=x^n+y^n? If not, give a counterexample.

Answer 1

try replacing n with a variable. What variable makes it true? What variable makes it not true? I always start with 2.

Answer 2

Right. Try something simple, such as n = 2, and x = y = 1.

Answer 3

"Freshman's dream"

Answer 4

If I replace n with 2, I come out with \[x^2+xy+y^2 = x^2 + y^2 \]. That is not true?

Answer 5

No. Suppose x=y=1. Are the two sides equal?

Answer 6

Do you mean x+y?

Answer 7

and btw,
\[ (x+y)^2 = x^2 + 2xy + y^2 \]

No: I mean x = 1 and y =1.

Answer 8

http://en.wikipedia.org/wiki/Freshman%27s_dream

Answer 9

But without expanding it out:
if x = 1 and y = 1, then
x+y = 2
and thus
\[ (x+y)^2 = 2^2 = 4 \]

But
\[ x^2 + y^2 = 1^2 + 1^2 = 1 + 1 = 2 \]

Hence for n = 2 and x=y=1, is it true that
\[ (x+y)^n = x^2+y^n \]
?

Answer 10

**correction: \[(x+y)^n = x^n + y^n \]

Answer 11

No it's not true. Did we use the Power of a Power Postulate?

Answer 12

No. Nothing so fancy was necessary.

Answer 13

It asks me to list what Postulate was implemented, and what do they mean by counterexample in this case?

Answer 14

You'll have to figure out what postulates were used here, given the list you have, as there's no standard list I can use to answer that for you.

A counter-example of a proposition in mathematics (or logic more generally) is an explicit example of when the result does not hold and hence the proposition is not true in general.

Let P be the following proposition:
For all real numbers x and y and all integers n
\[ (x+y)^n = x^n + y^n \ \ \ \ -- (*) \]

A counter-example to this proposition P is x=y=1 and n=2, because for those values of the variables, the equation (*) does not hold. Hence the proposition P is false in general. I.e., it is NOT the case that for all real numbers x and y, and all integers n that
\[ (x+y)^n = x^n + y^n \]

Answer 15

That clears things up a lot! Thank you very very much.

Answer 16

Sure thing