Mathematics, 01.09.2020 17:01 finldani4

Seven is part of all of the following sets of numbers except O irrational numbers
O integers
O rational numbers
natural numbers

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Answer from: thelonewolf5020

Seven isn't part of the set of irrational numbers.

Step-by-step explanation:

Seven and the Set of all Natural Numbers

Start with the smallest set among the choices. The set of all natural numbers, $\mathbb{N}$ , starts with (or , for some people.) A number is in $\mathbb{N}\!\!\!$ (write $n \in \mathbb{N}$ ) if and only if (n - 1) is in $\mathbb{N}\!$ . Conversely, if $n\!$ is indeed in $\mathbb{N}\!\!\!\!\!$ , then (n + 1) would also be in $\mathbb{N}\!\!\!\!\!\!\!$ . For :

$1 \in \mathbb{N} \implies 2 \in \mathbb{N}$ .

$\vdots$ .

$6 \in \mathbb{N} \implies 7 \in \mathbb{N}$ .

Therefore, is indeed in the set of all natural numbers.

Seven and the Set of all Integers

Similarly, a number is in the set of integers, $\mathbb{Z}$ , if and only if either (n - 1) or (n + 1) is (or both are) in $\mathbb{Z}\!\!$ .

Conversely, if a number is in $\mathbb{Z}$ , then both (n - 1) and (n + 1) will be in $\mathbb{Z}\!$ .

It can be shown in a similar iterative way that $7 \in \mathbb{Z}$ .

Alternatively, consider the fact that the set of all natural numbers, $\mathbb{N}$ , is a subset of the set of all integers, $\mathbb{Z}$ . Therefore, $7 \in \mathbb{N}$ implies that $7 \in \mathbb{Z}$ .

Seven and the Set of all Rational Numbers

A number is a member of the set of all rational numbers $\mathbb{Q}$ if and only if there exists two integers and such that:

$\displaystyle m = \frac{p}{q}$ .

and are both integers. If p = 7 and q = 1 , then $\displaystyle 7 = \frac{7}{1} = \frac{p}{q}$ . Hence,

Alternatively, note that the set of all integers, $\mathbb{Z}$ , is a subset of the set of all rational numbers, $\mathbb{Q}$ . Therefore, the fact that $7 \in \mathbb{Z}$ would imply that $7 \in \mathbb{Q}$ .

Seven and the Set of all Irrational Numbers

A number is in the set of all irrational numbers if and only if:

this number is in the set of all real numbers, andthis number is not in the set of all rational numbers. (Hence "irrational.")

Therefore, the fact that is a rational number implies that it is not an irrational number.

Answer from: jjimenez0276

irrational numbers

Step-by-step explanation:

Answer from: Quest

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

Answer from: Quest

nswer: a y-axis

step-by-step explanation:

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Seven is part of all of the following sets of numbers except O irrational numbers
O integers<...

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