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NeX [460]
3 years ago
9

X=1, X^2=X, X^2-1=X-1, (X+1)(X-1)=X-1, X+1=1, X=0 Where is the mastic?

Mathematics
1 answer:
Alchen [17]3 years ago
5 0

Answer:

You are technically dividing by zero, it breaks math.

In the step (X+1)(X-1)=X-1 → X+1=1.

You divide (X+1)(X-1)=X-1 by (x-1) on both sides to get:

(x+1)(x-1) / (x-1) = (x-1) / (x-1).

Since x = 1,

(x + 1)(x - 1) / (x - 1) = (x - 1) / (x - 1)

becomes (2)(0) / <u>0</u> = 0 / <u>0</u>.

And when you divide by zero, it's undefined since you cannot multiply a number by zero and get a number other than zero, it's undefinable.

This is where limits come in, when you divide by a variable, you have to assumed that is it ≠ 0.

You know that 1 ≠ 2, and 0 ≠ 1, and so on.

As dividing by zero approaches infinity,

1/0.1 = 10, 1/0.01 = 100, 1/0.001 = 1000, 1/0.00..1 = 1000.. → 1/0 → infinity.

1/-0.1 = -10, 1/-0.01 = -100, 1/-0.001 = -1000,

1/-0.00..1 = -1000.. → 1/0 → -infinity.

This suggests that dividing by zero should be infinity.

Therefore infinity = -infinity, Wrong.

When you are dividing, you are really asking yourself, how many times does this number fit into the other number.

1 cannot fit into 0, there isn't anything to fit into. This can also be thought of as:

1 = 0 + 0 + 0 + 0 + 0 + ......

0 + 0 = 0, so 1 ≠ 0.

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