No. For example, in real numbers, the square root of negative numbers are not defined.
No because all integers are rational numbers
Yes. Every integer is a rational number. Every rational number is a real number. Every real number is a complex number. The complex numbers include all real numbers and all real numbers multiplied by the imaginary number i=sqrt(-1) and all the sums of these.
Not at all. Every whole number is rational.
Numbers are split into real and imaginary. Rational numbers are under the category: Real. Therefore all rational numbers are real. An irrational number is also real, but can not be expressed as a fraction.
No if the denominators cancel each other out there is no asymptote
No.The equation x/(x^2 + 1) does not have a vertical asymptote.
Answer: no [but open to debate] ((x-1)(x-2)(x+2))/(x-3) (x^2-3x+2)/(x-2)(x+2) Asymptote missing, graph it, there is no Asymptote because the (x-2)(x+2) can be factored out. yes
Many functions actually don't have these asymptotes. For example, every polynomial function of degree at least 1 has no horizontal asymptotes. Instead of leveling off, the y-values simply increase or decrease without bound as x heads further to the left or to the right.
Every function has a vertical asymptote at every values that don't belong to the domain of the function. After you find those values you have to study the value of the limit in that point and if the result is infinite, then you have an vertical asymptote in that value
Yes. There is an injective function from rational numbers to positive rational numbers*. Every positive rational number can be written in lowest terms as a/b, so there is an injective function from positive rationals to pairs of positive integers. The function f(a,b) = a^2 + 2ab + b^2 + a + 3b maps maps every pair of positive integers (a,b) to a unique integer. So there is an injective function from rationals to integers. Since every integer is rational, the identity function is an injective function from integers to rationals. Then By the Cantor-Schroder-Bernstein theorem, there is a bijective function from rationals to integers, so the rationals are countably infinite. *This is left as an exercise for the reader.
Yes, because for every X input there are multiple y values.
Every integer is a rational number.
Horizontal line test is used for the determination of a function,if the horizontal line passes through one point of the given graph then it is a function and if it passes through more than one point then it will not a function. * * * * * No! It is a vertical line test. Consider the graph of y = sin(x): a horizontal line line will cross it twice in every 360 degrees! Convince me that y = sin(x) is not a function.
If for every point on the horizontal axis, the graph has one and only one point corresponding to the vertical axis; then it represents a function. Functions can not have discontinuities along the horizontal axis. Functions must return unambiguous deterministic results.
Find All Possible Roots/Zeros Using the Rational Roots Test f(x)=x^4-81 ... If a polynomial function has integer coefficients, then every rational zero will ...
It is a function because for every point on the horizontal axis, the parabola identified one and only one point in the vertical direction.