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DanielleElmas [232]
3 years ago
7

Use complete sentences to describe why √-1 ≠ -√1

Mathematics
2 answers:
tekilochka [14]3 years ago
5 0

Well let's say that to compare these two numbers, we have to start with the definition first.

<u>D</u><u>e</u><u>f</u><u>i</u><u>n</u><u>i</u><u>t</u><u>i</u><u>o</u><u>n</u>

\displaystyle \large{ {y}^{2}  = x} \\  \displaystyle \large{ y =  \pm  \sqrt{x} }

Looks like we can use any x-values right? Nope.

The value of x only applies to any positive real numbers for one reason.

As we know, any numbers time itself will result in positive. No matter the negative or positive.

<u>D</u><u>e</u><u>f</u><u>i</u><u>n</u><u>i</u><u>t</u><u>i</u><u>o</u><u>n</u><u> </u><u>I</u><u>I</u>

\displaystyle \large{  {a}^{2}  = a \times a =  |b| }

Where b is the result from a×a. Let's see an example.

<u>E</u><u>x</u><u>a</u><u>m</u><u>p</u><u>l</u><u>e</u><u>s</u>

\displaystyle \large{  {2}^{2}  = 2 \times 2 = 4} \\  \displaystyle \large{  {( - 2)}^{2}  = ( - 2) \times ( - 2) =  | - 4|  = 4}

So basically, their counterpart or opposite still gives same value.

Then you may have a question, where does √-1 come from?

It comes from this equation:

\displaystyle \large{   {y}^{2}  =  - 1}

When we solve the quadratic equation in this like form, we square both sides to get rid of the square.

\displaystyle \large{   \sqrt{ {y}^{2} } =   \sqrt{ - 1}  }

Then where does plus-minus come from? It comes from one of Absolute Value propety.

<u>A</u><u>b</u><u>s</u><u>o</u><u>l</u><u>u</u><u>t</u><u>e</u><u> </u><u>V</u><u>a</u><u>l</u><u>u</u><u>e</u><u> </u><u>P</u><u>r</u><u>o</u><u>p</u><u>e</u><u>r</u><u>t</u><u>y</u><u> </u><u>I</u>

\displaystyle \large{  \sqrt{ {x}^{2}  } =  |x|  }

Solving absolute value always gives the plus-minus. Therefore...

\displaystyle \large{  y =   \pm \sqrt{ - 1}  }

Then we have the square root of -1 in negative and positive. But something is not right.

As I said, any numbers time itself of numbers squared will only result in positive. So how does the equation of y^2 = -1 make sense? Simple, it doesn't.

Because why would any numbers squared result in negative? Therefore, √-1 does not exist in a real number system.

Then we have another number which is -√1. This one is simple.

It is one of the solution from the equation y^2 = 1.

\displaystyle \large{   {y}^{2}  = 1} \\  \displaystyle \large{    \sqrt{ {y}^{2} }  =  \sqrt{1} } \\  \displaystyle \large{  y  =  \pm  \sqrt{1} }

We ignore the +√1 but focus on -√1 instead. Of course, we know that numbers squared itself will result in positive. Since 1 is positive then we can say that these solutions exist in real number.

<u>C</u><u>o</u><u>n</u><u>c</u><u>l</u><u>u</u><u>s</u><u>i</u><u>o</u><u>n</u>

So what is the different? The different between two numbers is that √-1 does not exist in a real number system since any squared numbers only result in positive while -√1 is one of the solution from y^2 = 1 and exists in a real number system.

ivolga24 [154]3 years ago
4 0

Answer:

Therefore \sqrt{1} \neq -\sqrt{1}

Step-by-step explanation:

\sqrt{-1}  \\ cannot be calculated as no two identical numbers that are multiplid could give -1. However - \sqrt{1} means -1 x \sqrt{1} = -1 x 1 = -1 . Therefore \sqrt{1} \neq -\sqrt{1}

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topjm [15]

Answer:

<h2>1/7</h2>

Step-by-step explanation:

If I choose a number from the integers 1 to 25, the total number of integers I can pick is the total outcome which is 25. n(U) = 25

Let the probability that the number chosen at random is a multiple of  6 be P(A) and the probability that the number chosen at random is is larger than 18 be P(B)

P(A) = P(multiple of 6)

P(B) = P(number larger than 18)

A = {6, 12, 18, 24}

B = {19, 20, 21, 22, 23, 24, 25}

The conditional probability that the number is a multiple of 6 (including 6) given that it is larger than 18 is expressed as P(A|B).

P(A|B) = P(A∩B)/P(B)

Since probability = expected outcome/total outcome

A∩B = {24}

n(A∩B) = 1

P(A∩B) = n(A∩B)/n(U)

P(A∩B) = 1/25

Given B = {19, 20, 21, 22, 23, 24, 25}.

n(B) = 7

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p(B) = 7/25

Since P(A|B) = P(A∩B)/P(B)

P(A|B) = (1/25)/(7/24)

P(A|B) = 1/25*25/7

P(A|B) = 1/7

<em></em>

<em>Hence the conditional probability that the number is a multiple of 6 (including 6) given that it is larger than 18 is 1/7</em>

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a spherical ball with a volume of 7776pi cm cubed is packaged in a box that is in the shape of a cube. the edge length of the bo
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4 0
3 years ago
Use these functions to answer the questions.
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Part A
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     \left(x+3\right)^2=1

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     x+3=1

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The x-intercept is (-2,0). 

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The y-intercept is (0, 8) 

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Part B.
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4 0
4 years ago
What is the answer to 5/8 . 2/7 and what does the dot stand for the dot is actually a little higher
Illusion [34]
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= 5/28

The final answer is 5/28~
6 0
4 years ago
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