Natural numbers, integers, etc. What are the definitions of these numbers? I immediately went to senior one, suddenly found some primary school nouns do not understand 1: Integers 2: Natural number 3: Rational and irrational numbers 4: Real and imaginary numbers 5: Plural 6: Common divisor and common multiple 7: Even number and odd number (this can not be explained)! I hope you can tell me what the definition of these numbers is? For example, it would be better! I also forgot which * * numbers are still available in primary schools! If there are any, please answer!

Natural numbers, integers, etc. What are the definitions of these numbers? I immediately went to senior one, suddenly found some primary school nouns do not understand 1: Integers 2: Natural number 3: Rational and irrational numbers 4: Real and imaginary numbers 5: Plural 6: Common divisor and common multiple 7: Even number and odd number (this can not be explained)! I hope you can tell me what the definition of these numbers is? For example, it would be better! I also forgot which * * numbers are still available in primary schools! If there are any, please answer!

Natural numbers above 1.0, including 0
2.1、2、3……
3. All are rational numbers except infinite acyclic decimals
4. Real numbers include rational numbers and irrational numbers, imaginary numbers also include non pure imaginary numbers and pure imaginary numbers. The form of non pure imaginary numbers is a + bi, while the form of pure imaginary numbers is Bi, where I is the unit
5. The complex number is composed of real number and imaginary number
For example, 2 and 4, 2 is a common divisor and 4 is a common multiple

All kinds of number concepts, such as natural numbers, integers and so on, please be more complete

All integers are natural numbers, right?
Is 0 a natural number

All integers are natural numbers, right?
Wrong. Integers include positive integers, 0 and negative integers, while natural numbers only include positive integers and 0
0 is a natural number

40% of a number is 20 less than one and a fifth of it. What's the number

40% of a number is 20 less than one and a fifth of it. What's the number
Let this number be X
1 and 1 / 5x-40% x = 20
0.8x=20
x=25

Six cups are arranged in a row, three on the left are filled with water, and three on the right are empty,
That's what it says in the book

Hello
Six cups are arranged in a row, three on the left are filled with water, and three on the right are empty
We set the water cup from left to right as 1.2.3 and the right three empty cups from left to right as 4.5.6
It is shown as follows:
1 (with water) 2 (with water) 3 (with water) 4 (without water) 5 (without water) 6 (without water)
If you think about it carefully, you can figure it out: just change "2 (with water)" and "5 (without water)"
1 (with water) 5 (without water) 3 (with water) 4 (without water) 2 (with water) 6 (without water)
You see right?

(2014. The first mock exam in Hohhot) {an}, known for any positive integer n, a1+a2+a3+... +An = 2N-1, then A12 + A22 + A32 + +An2 & nbsp; equals ()
A. （2n-1）2B. 13(2n−1)C. 13(4n−1)D. 4n-1

∵a1+a2+a3+… +an=2n-1… ①∴a1+a2+a3+… +an-1=2n-1-1… ② The results show that an = 2N-1, an2 = 22n-2, an2} is an equal ratio sequence with 1 as the first term and 4 as the common ratio, A12 + A22 + A32 + +An2 = 1 − 4n1 − 4 = 13 (4N − 1), so C

① 9-12 × 7 / 8 ／ 7 / 6 ② (3 / 4 ／ 7 / 8-5 / 14) ／ 7 / 16 simple calculation, drag calculation

1、9-12*7/8*6/7=9-9=0
2、（3/4*8/7-5/14）*16/7=（6/7-5/14）*16/7=1/2*16/7=8/7

Given the set a = {2, A2 + 1, a2-a}, B = {0,7, a2-a-5,2-a}, if 5 ∈ a, find the set B

This kind of problem is not difficult, as long as you know that the set elements are different
5 ∈ a, then a ^ 2 + 1 = 5, a = 2, or - 2
Substituting a ^ 2-A = 2 (rounding off when a = 2), or 4 (a = - 2), the,
Then set a = {2,5,6}
Substituting a = - 2 into B,
B={0,7,1,4}

In the ratio of 6:70.7:0.8 2 / 21:4 / 21, the ratio of 2 / 3:4 / 7 is the highest（
In the ratio of 6:70.7:0.8 2 / 21:4 / 21, the ratio of 2 / 3:4 / 7 is (), and the inward product of this ratio is ()

6：7

Let a be an M × n matrix and B be an n × s matrix. Given the rank (b) = n and ab = 0, it is proved that a = 0

Let R (b) = n, we know that the row vectors of B are linearly independent. Let the group of row vectors be: B1, B2,. BN, and divide B into blocks by row,
(B 'denotes transpose of B)
B = (B1, B2,..., BN)
Let a = [a (ij)] I = 1,2,. M, j = 1,2,. N
In this way, AB still has a matrix C: which is partitioned by row
AB=C=[C1,C2,.,Cm]'.
Where CK = a (K1) B1 + a (K2) B2 + a (K3) B3 +. + a (KN) BN. (k = 1,2,..., m)
According to the assumption: ab = 0, that is, CK = 0,
Since B1, B2,... BN are linearly independent, it is deduced that a (K1) = a (K2) = a (K3) =. = a (KN) = 0
(k=1,2,3,.,m)
That is to say: a = 0