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Mathematicians
Often, to completely understand the importance of a mathematical theory, it's essential to know about the person who developed it. This category is where all questions about historically significant mathematicians should be asked.
6,570 Questions
If john is 55 inches and shania is 1.1 times taller than him how tall is shania?
Shania is 60.5 inches.
The remainder is 0.
If A has a remainder of 1 when divided by 3, then A = 3m + 1 for some integer m
If B has a remainder of 2 when divided by 3, then B = 3n + 1 for some integer n
→ A + B = (3m + 1) + (3n + 2)
= 3m + 3n + 1 + 2
= 3m + 3n + 3
= 3(m + n + 1)
= 3k where k = m + n + 1 and is an integer
→ A + B = 3k + 0
→ remainder when A + B divided by 3 is 0
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From this, you may be able to see that:
- if A when divided by C has remainder Ra; and
- if B when divided by C has remainder Rb; then
- (A + B) divided by C will have remainder equal to the remainder of (Ra + Rb) divided by C
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A can be 0 or 9.
A + b = 0, 9, 18
(9,0)(8,1)(7,2)(6,3)(5,4)(4,5)(3,6)(2,7)(1,8)(0,9)(0,0)(9,9)
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To be divisible by 9, the sum 6 + 3 + 1 + 8 + 7 + 2 + A + B = 27 + A + B must be divisible by 9
→ A + B must be a multiple of 9.
→ (A, B) must be (0, 0), (0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0)
Note that (0, 0) must not be forgotten as 27 + 0 + 0 = 27 = 3 × 9.
What irrational number has the decimal representation that begins 0.232323232?
23/99, which is rational.
What fraction of whole numbers is equal to 17100 and that has a denominator that is a power of 10?
17100 = 17100/1
= 17100/1 × 10/10 = 171000/10
= 17100/1 × 10²/10² = 1710000/100
= 17100/1 × 10³/10³ = 17100000/1000
...
= 17100/1 × 10ⁿ/10ⁿ
There is no upper limit to this sequence: n = 0, 1, 2, 3, ....
These numbers can be put in a one-to-one relationship with the counting numbers 1, 2, 3, ...
The counting numbers can also be put in a one-to-one relationship with the whole numbers
Therefore the sequence of fractions can be put in a one-to-one relationship with the whole numbers.
Therefore the fraction of whole numbers which are in the sequence is 1.
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Let's try another approach:
As 10 to some power is positive, there is no way we can change a negative number to make it positive by dividing by 10 to some power, therefore the solution numbers are all positive.
As negative whole numbers make up half of the whole numbers, the fraction of whole numbers which are equivalent to 17100 when divided by a power of 10 is at most ½.
The fraction of the ½ of the whole numbers which are positive can now be considered:
After the nth whole number has been found which matches the criteria that is it equivalent to 17100 when it is divided by a power of 10, in total there are n numbers matching out of a total of 17100 × 10ⁿ⁻³ numbers (the value of the nth number); thus:
The first number which meets the criteria is 171/10⁻² → the fraction is 1/171
The second number to meet the criteria is 1710/10⁻¹ → the fraction is 2/1710
The third number to meet the criteria is 17100/10⁰ → the fraction is 3/17100
The fourth number to match the criteria is 171000/10 → the fraction is 4/171000
The fifth number to match the criteria is 1710000/10²→ the fraction is 5/1710000
→ For the nth match, the fraction is: n/(17100 × 10ⁿ⁻³) = (1/17100) × n/10ⁿ⁻³
This gives us a sequence of fractions:
1/17100 × 1/10⁻², 1/17100 × 2/10⁻¹, 1/17100 × 3/10⁰, 1/17100 × 4/10¹, 1/17100 × 5/10², ...
= 100000/17100000, 20000/17100000, 3000/17100000, 400/17100000, 35/17100000, 6/17100000, ....
Consider terms n and n+1:
U{n} = (1/17100) × n/10ⁿ⁻³ = (1/17100) × 10 × n/10ⁿ⁻²
U{n+1} = (1/17100) × (n+1)/10ⁿ⁻²
U{n} - U{n+1} = (1/17100) × 10 × n/10ⁿ⁻² - (1/17100) × (n+1)/10ⁿ⁻²
= (10n - (n+1))/(17100 × 10ⁿ⁻²)
= (9n - 1)/(17100 × 10ⁿ⁻²)
As n ≥ 1,
9n - 1 ≥ 9×1 - 1 = 8
→ term n - term n+1 ≥ 8 > 0
→ term n is larger than term n+1
for all n ≥ 1
Each of these terms is less than 1, and each term of the sequence is smaller than the previous one and so as n increases the value of the each term (the fraction of whole numbers which meet the criteria) tends towards 0.
→ The fraction of all whole numbers which equate to 17100 when divided by a power of 10 is as near enough to zero as make no odds.
ie the fraction is so small it is effectively none of the whole numbers.
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This is a problem of dealing with the infinite.
What Two angle measures of this triangle are 60 and deg. Which type of triangle is it?
An equilateral triangle would fit the given description
let the number of bottles in the first group be B
Then
- the 2nd group has 1st + 20 = B + 20 bottles;
- the 3rd group has 2nd × 2 = 2 × (B + 20) = 2B + 40 bottles;
- the 4th group has 3rd + 10 = 2B + 40 + 10 = 2B + 50 bottles.
The total number of bottles is the sum of the groups which is:
bottles = B + B + 20 + 2B + 40 + 2B + 50 = 6B + 110
But there are 1700 bottles, we can solve for B:
6B + 110 = 1700
→ 6B = 1590
→ B = 265
→ the groups contain:
- B = 265 bottles
- B + 20 = 265 + 20 = 285 bottles
- 2B + 40 = 530 + 40 = 570 bottles
- 2B + 50 = 530 + 50 = 580 bottles
The groups contain 265, 285, 570 and 580 bottles respectively.
What is the surface area of a sphere with a diameter of 30cm?
Surface area of sphere: 4*pi*15^2 = 900*pi square cm
What is 0.424242 written as an fraction?
0.424242 is a fraction. It is a fraction in decimal form rather than in the form of a ratio. However, that does not stop it being a fraction. Its equivalent, in rational form, is 424242/1000000. You can simplify this rational fraction if required.
It is 7:00 am.
They are three terms of an expression that can be simplified to: 3p -31
How do you show proof by contradiction That 3 to the power of x equals 2 Is Irrational?
Using "^" for powers.
3^x = 2. Assume that x is a rational number, and call it p/q. So:3^(p/q) = 2
Raise both parts to the power "q", and you get:
(3^(p/q))^q = 2^q
or:
3^p = 2^q
You must now raise 3 to an integer power "p", and 2 to an integer power "q", and get the same result. This is clearly impossible, since 3 and 2 have no common factors.
Put it another way: 3 raised to any power will always be ODD, 2 raised to any power will always be EVEN. The two therefore can't be equal.
This contradicts the initial assumption, that p and q are integers, i.e., that the solution is an integer number, p/q.
