Google Math

spazntwitch

spazntwitch

It's like déj- vu again
Joined
Jan 7, 2003
Messages
9,355
Reaction score
970
Points
0
Just did a calculation with Google and discovered that 1 / 0 = 0. :banghead:

Must be the new Google math!

To try this amazing and neato trick at home, go to Google and type 1/0= in the search box and press the Enter key on the keyboard.
 
Brado said:
It is on most calculators.
Click to expand...

not really all calculators i have return either undefined or a non zero value or error message.
 
spazntwitch said:
Just did a calculation with Google and discovered that 1 / 0 = 0. :banghead:

Must be the new Google math!
Click to expand...

Is it wrong? Doesn't 1 divided by 0 = 0 ???
 
iluvdeals said:
Is it wrong? Doesn't 1 divided by 0 = 0 ???
Click to expand...
It should be undefined. You cannot divide something by nothing.
 
spazntwitch said:
It should be undefined. You cannot divide something by nothing.
Click to expand...

Gotcha!

Hmmm... You must have been really bored to figure that out. ;)
 
actuallt its infinity because even 0 has a value , its just defined , so the answer should be infinity , its one of the first things you learn in math
 
yup, the answer is "infinity"...or in programming , infinite loop





ps, nice new smiley icons!! ;)
 
A friend of mine posted this on another website, and the answer eludes me, too. Can one of you spoofee geniuses help?

-----------------------
I used to be pretty good at math, but I was just looking a sample question from the SAT, and for the life of me, I can't figure out the answer, even when it's explained. Here's the question:

1. |4x - 7| = 5
|3 - 8x| = 1

What value of x satisfies both of the equations above?

And here's the answer and the explanation:

Since |4x - 7| = 5
the value of |4x - 7| is either 5 or -5

4x-7=5 or 4x - 7 = -5
4x=12 4x = 2
x=3 x = 1/2

The two values of x that satisfy the first equation are 3 and 1/2

Since 3-8x=1 the value of 3-8x is either 1 or -1

3 - 8x =1 or 3 - 8x = -1
8x + 2 8x = 4
x = 1/4 x = 1/2

The two values of x that satisfy the second equation are 1/4 and 1/2

You are asked to find the value of x that satisfies both equations. That value is 1/2
------------------

Here are the things that have me confused:

(1) If |4x - 7| = 5, then how can the value of |4x - 7| be either 5 or -5?

(2) Similarly, if 3-8x=1, then how can the value of 3-8x be either 1 or -1?


(I edited this to correct a typo in the original posting.)
 
Last edited:
Think of it this way:
|5| = 5
|-5| = 5

and similarly:
|1| = 1
|-1| = 1
 
The |x| marks denote "absolute value". The absolute value is the number regardless of its sign, ie regardless if it's positive or negative. The number will be reported as the positive number.

Absolute values are important... ie why the Kelvin and Rankine temperature scales were made.
 
dehawk666 said:
actuallt its infinity because even 0 has a value , its just defined , so the answer should be infinity , its one of the first things you learn in math
Click to expand...

Actually, it's not infinity, because infinity isn't a real number.

If you take the limit of the series:
1/(1/2); 1/(1/3); 1(1/4); 1/(1/5); 1/(1/6); 1/(1/7); 1/(1/8).....

You see the demominator is approaching zero... so you'd eventually hit 1/0.
The series above progresses: 2; 3; 4; 5; 6; 7; 8.... infinity.

If you take a look at this series:
1/(-1/2); 1/(-1/3); 1(-1/4); 1/(-1/5); 1/(-1/6); 1/(-1/7); 1/(-1/8).....

Notice the denomintor is also approaching zero... so you'd also eventually hit 1/0.
Except this series above progresses: -2; -3; -4; -5; -6; -7; -8.... negative infinity.

How can 1/0 be both infinity and negative infinity? It can't; and therefore it is considered UNdefined.
 
Approaching 0 but will never hit 0 because the denominator will approach infinity . I may be wrond on this one but the denominator will never hit 0
 
The denominator never approaches infinity. The demominator is getting closer and closer to zero each time. Which means the actual number:

1/(getting closer and closer to zero)

...is actually getting closer and closer to infinity... and also negative infinity (as seen in my second example); wherin the problem lies. How can the value 1/0 approach two different numbers? Both infinity and negative infinity. It can't. Which is why the answer is undefined, not infinity; not negative infinity.

For another math, uh, mind-bender: Mathmatically, .999999999... = 1. Yeah, equals, at least by mathmatical definitions.
 
Talking about .9999999999999999999....= 1, it actually doesnt, but here is some "proof" that it does.
If X=.99999999999...
Then 10X=9.9999999999999...
Then 10X-X= 9X
And 9.99999999..... -.99999999999=9
9X=9
X=1

It's kinda confusing, but if you think about it, there is a ".0000.........1" difference
 
hmm another thing is that the .000... 1 is from
.999 infinite... but when u multiply it by 10, its moves decimal over 1 place... .999 infinite - 1. (technically)
therefore when you do 10x-x = 9x.. 9.999999... - .999999 (remember ur missing a place) = 9.0000000....1

this would i work if the numbers are infinite... but will not work if the number was finite....
 
Too much...brain going to explode...

lol
 
There is no "0.000000.....1" difference because you'll never get to that "1." It's more like a "0.000000...." difference conceptually.

Two numbers are the same if you can't find any other number inbetween the two. You can't find a number between 0.9999999... and 1.

But I'd like to see the 1=0 proof. It pretty much fails because there's always a "divide by zero" step.
 
You must log in or register to reply here.
Back
Top