(A real life problem:
A millionaire passed away and left down 6 millions. He had 2 wives, and each wife had c children.
In his will, the wealth would be divided equally among all his children. So, how much would each of the children receive?
Naturally we will come out with this:
6millions ÷ 2c, since the total number of children = 2*c=2c
Now,
If c=(1 girl + 2 boys),
option (1) => 6millions ÷ 2(1+2) = 6millions * 1/2*1/3 = 1 million
option (2) => 6millions ÷ 2(1+2) = 6millions ÷ 2*3 = 9 millions
by option 1, each child would get 1 million,
by option 2, each child would get 9 millions.
Which one is correct?
(From one comment below, someone says it should be 6millions ÷ [2(1+2)] instead.
6millions ÷ [2(1+2)] is true, but since [2*(1+2)] = [2(1+2)] = 2(1+2), we may just drop the parenthesis. The detail is included in the proof below.
)
6÷2(1+2) = 1
and
30÷2(2+3)÷5=0.6.
But... How could this confuse so many people? And there are so many people arguing that the correct answers should be 9 for the first problem and 15 for the second problem. Even the news and some professionals argue that the correct answers are 9 and 15 respectively.
However, I am quite sure most students from PRC will get the correct answers, which are 1 and 0.6 respectively.
In fact, there are many students gave the correct answers, but they were told that they were wrong. What the heck?!!
*[To my dear readers, if you have forgotten some properties of function, please read the 'Extra' at the end of the post before you proceed. Thanks =). ]
Why is 6÷2(1+2) = 1, not 9?
Let c=1+2. Compute 6÷2c.
Then 6÷2c = 6*1/2c = 6/2c = 3/c .
( How could they get 3c??? Sigh... they do it in this way: 6÷2c = 6÷2*c = 3c .
Well, assume to the contrary that this is true, then
Well, assume to the contrary that this is true, then
6÷2c = 6÷2*c = 6*1/2*c = 6*c*1/2 = 6c/2 = 6c ÷ 2, which is a contradiction.
So, 6÷2c ≠ 6÷2*c for c≠1).
Why? Let a,b,c be any real numbers.
Then a÷b*c = a*1/b*c = ac/b ≠ a/bc = a*1/bc = a÷bc , for a≠b, and c≠1.
(Note: bc is a number, while b*c is the product of two numbers b and c, and b*c=bc.
If we write b=2, c=3, then 2*3 = 6, so bc=6 is a number, while 2*3 is the product of two numbers 2 and 3.
If we write a=6, b=2, c=3, then a÷b*c = 9 , but a÷bc = 1.
If we assume to the contrary that a÷bc=9, then from the previous argument, a÷bc=6÷6=9, which is a contradiction.)
(From one comment below, someone is confused when I assert b*c=bc but on the other hand I said 6÷2c ≠ 6÷2*c. I know this is trivial and I didn't want to be so nagging on this part. Anyway, since some people might not know, let me give a little bit explanation. As from the example above, 2*3 = 6, then 6÷(2*3) = 6÷6. Generally for b*c=bc, then a÷(b*c)=a÷bc. But, a÷b*c≠a÷bc if c≠1 and a≠b.
Still not convinced? Well, if alternatively, we write a÷b², we know that (b*b)=bb=b², how do you interpret a÷bb? )
Hence 6÷2(1+2) = 1,
but 6÷2*(1+2) = 9.
Most students never learn the proof above, including myself. Why?
Because it is so clear that we don't necessarily need to go through the proof to get 6÷2(1+2) = 1.
Sigh...
While for the second problem, 30÷2(2+3)÷5, it follows the similar argument that the answer is 0.6.
In fact, any qualified scientific calculators would give us the correct answers.
All the news below are bull shits!!!
Even the math puzzle in the end of the news below was incomplete/not the real one we learned, which again, misled the students.
The puzzle given by the teacher was:
3 ppl went to a hotel, the initial cost was $10 per person. But they were given 10% discount, which means the total cost = 3*9=27. However the waiter put $2 into his own pocket. Then $27+$2 =$29, where is the missing $1? (Is it a puzzle? Nope, it is so obvious that if you paid 27, while the waiter kept $2, the resulting payment the hotel received was $25. Who would go and calculate $27+$2 ?)
A more accurate puzzle should be:
3 ppl went to a hotel and rent a bedroom, the room initially cost $30. Each of them gave $10 to a waiter who helped them made the payment. But the casher told the waiter that the room cost $25 only for that day and returned $5 to the waiter in coins. But the waiter realized that he could not divide the change equally btw the 3 ppl. So he decided to put $2 into his own pocket and give $1 to each of the 3 ppl.
Now each of the 3 ppl had been given a dollar back, so each of them paid $9. Then 3*$9 = $27.
The waiter had $2 in his pocket. So, $2+ $27 = $29. The 3 ppl originally handed over $30, where is the missing dollar?
The following news is even more ridiculous. The higher quality calculator indeed gives the correct answer, while the other cheaper calculator gives the wrong one. But the broadcaster told a different story.
Some people argue that by PEMDAS rule, the answers should be 9 for the 1st problem.
Note:
P - parentheses
E - exponents
M - multiplication
D - division
A - addition
S - subtraction
E - exponents
M - multiplication
D - division
A - addition
S - subtraction
PEMDAS is true, but using this rule without following other rules is not the right way to do mathematics.
(Extra:
When we learn function, we ordinarily write xy=z for x*y=z, where x and y are independent variables while z is the dependent variable.
We can see that for every real number x and y, x*y = xy. So, for each x and y, z=xy.
Writing xy=z is better than writing x*y=z because it is unambiguous and convenient.
Let say we are given the values of x and y, and we now want to compute a/z, where a is some constant.
By convention, for each real number x and y, we compute a/xy to get a/z.
We don't write a/x*y.
This is why when we write 6/2(1+2), we don't write 6/2*(1+2), they are different from each other.)
