Question #201710

Define a binary relation P from R to R as follows: for all real numbers x and y, (x,y)∈P⇔x=y^2. Is P a function? Explain.


1
Expert's answer
2021-06-02T11:05:23-0400

Taking into account that 4=224=2^2 and 4=(2)24=(-2)^2, we conclude that (4,2)P(4,2)\in P and (4,2)P(4,-2)\in P. Therefore, the element x=4Rx=4\in \mathbb R corresponds to two elements y=2Ry=2\in\mathbb R and y=2Ry=-2\in\mathbb R, and hence P:RRP:\mathbb R\to\mathbb R is not a function y=f(x).y=f(x).


On the other hand, for each element yRy\in\mathbb R there exists a unique element x=y2Rx=y^2\in\mathbb R. Consequently, P:RR,P(y)=y2,P:\mathbb R\to \mathbb R, P(y)=y^2, is a function of variable y.y.


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Comments

Assignment Expert
15.07.21, 21:19

Dear Amar, please use the panel for submitting a new question.


Umair
03.06.21, 06:26

I am very thankful for your response. Good bless you.

Amar
02.06.21, 12:43

Subject: Affine and Euclidean Geometry Question: Generalized geometrical incidence

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