# Math errors

1. Aug 28, 2006

### lo2

Well when calculating Math I sometimes make errors (we all do of course) but some of them are because I do not know the 'laws of math' (do not know what you call it in English). An examlpe.

$$\ln{2}=x^2+x+k \Leftrightarrow e^{\ln{2}}=e^{x^2}+e^x+e^k \Leftrightarrow 2=e^{x^2}+e^x+e^k$$

But that is wrong it should be.

$$\ln{2}=x^2+x+k \Leftrightarrow e^{\ln{2}}=e^{x^2+x+k} \Leftrightarrow 2=e^{x^2+x+k}$$

Which is how it should be.

Now my request is could anyone please help me not making those errors again which is not because of the fact that Humans do make errors but because I simply did not the 'law' well enough.

2. Aug 28, 2006

### dmoravec

are you asking for a list of such laws, or a site of reference or just ways to try and remember the laws? The above one can be remembered by remembering that the exponential is not a linear function.

3. Aug 28, 2006

### lo2

I am requesting for something (what is it I will leave up to you) which can help me not making those errors.

4. Aug 28, 2006

### neutrino

Do a lot of problems. If you are stuck, you can post it here.

5. Aug 28, 2006

### chroot

Staff Emeritus
You're asking for something which does not exist. The only way to learn the rules is simply to use them until you get used to them.

- Warren

6. Aug 28, 2006

### t!m

I've always been a fan of the 'what you do to one side of the equation, you have to do to the other' school of thought when it comes to these kinds of mistakes. So in this case, imagine there are parentheses around each side of the equality, before taking the exponential. So (ln2) = (x^2 + x +k). Then take the exponential of the entire expression on both sides: e^(ln2) = e^(x^2 + x +k).

Now if your problem lies in understanding why e^(x^2 + x +k) does not equal e^(x^2) + e^x + e^k, then this is simply, as stated before, a matter of the linearity of the exponential function. A function, f, is said to be linear if f(c*x + d*y) = c*f(x) + d*f(y), where c and d are coefficients. Try and come up with a simple example which shows that e^x is not a linear funtion.

7. Aug 29, 2006

### HallsofIvy

I'm not sure what "laws" you mean since what you give here is, of course, not a "law".

What you are showing here is the very common "math error" of assuming that f(x+ y)= f(x)+ f(y) which is true only for the very simple function f(x)= ax, certainly not for the exponential function.

(Before you experts jump on me, I am aware that there exist very nasty, non-continuous functions that satisfy f(x+ y)= f(x)+ f(y) but I didn't think that was relevant here.)

8. Aug 29, 2006

### mtanti

It is funny how few teachers ever teach this 'balancing equations' correctly. What you are actually doing is using both sides of the equations as inputs to functions using the rule that...

if a = b then f(a) = f(b)

With that in mind you will make less mistakes because you have in mind that you are doing the operations on the whole a and the whole b.

EG x+2 = 2x
let function f be defined as f(a)=a-2x
therefore f(x+2) = f(2x) => (x+2)-2x = (2x)-2x => x+2-2x = 0

And that is what is really happening! How can you know that if a=b then f(a)=f(b)? Because if they are the same function and both inputs are the same then they have to give the same output! Otherwise the function is random and not consistent in pattern...

Notice, this can only apply if you are finding the value where a=b. Otherwise if they are not equal then you cannot assume that f(a)=f(b).

9. Aug 29, 2006

### d_leet

To be technical, this is only true if f is well defined.

10. Aug 30, 2006

### HallsofIvy

mtanti said that f was a function which covers that.

(Although I am inclined to disagree with mtanti that "few teachers ever teach this 'balancing equations' correctly." Perhaps I was lucky but all of the teachers I have ever had, going back to grade 9 algebra, emphasised that.)

11. Aug 30, 2006

### mtanti

Wow, maybe it's something having to do with my country Malta then... haha, my first algebra teacher said that when you have such an equation:
a = b + c
then to make b subject, you give c a kick and it is sent flying to the other side of the equals and the kick injures it so instead of a + you have a -. lol this mislead me to changing + to - even in multiplication and division balancing for a really long time! What a stupid/lazy teacher!

Last edited: Aug 30, 2006
12. Aug 30, 2006

### HallsofIvy

"Give c a kick and sent it flying"?? Couldn't you be arrested for that? No telling what injuries that poor little c might suffer. Seems to me a very violent way of doing mathematics.

There are a number of people who learn to "move" numbers or variables from one side of the equation to another (though not by "kicking" them!). In some cases those are people whom I KNOW were taught differently. I suspect most people find it less abstract to think about "moving" things- they can see the "x" starting on one side of the equation and ending up on the other- and translate what they are told into those terms.