Math?

Beanbaker · Math?

I was looking through some old stuff i found online and came across this little piece:

Now I question all that I have learned.......

Koku

I think there is an error in it actually. I am not sure, but I am sure there is something there that is wrong. The only thing that equals one besides one is .9999999continuem.

Delilah

Why would you substitute A with B? I think that's where the error is. I'm no math wiz but I don't think that's right.

Beanbaker

You can substitute A with B for A and B are both the same value, just like you could hypothetically substitute X with some number.

As for the .99999999 thing, I saw one just like that. I forgot the process, but it did end up with the .99999999, but this is just a simplified version I guess.

Koku

I think the major problem with this may just be the fact A=B. Two variables for the same thing seems illogical. I also think that some of the steps such as 4 are not corrcet.

As for the thing I pointed out it is because 1/9= .1111111111 so multiply it by nine to get the .9999999

Beanbaker

Well the two variables seem like the only way to do it... since they "vary" they could be anything I guess.

I'm also looking pretty mad on google for the process with the .99999, but only some stupid stuff comes up, like 2(x+y)=(x+y) or something like that, that just doesn't work anyways. The steps, however, they do seem legal to what I know. The factoring seems right.

Delilah

Thanks Beanbaker. I forgot about them being equal.

Koku

Well the big problem with this as I just had to look up is simply with the step when you go from (a-b)(a+b)=(a-b)(b) to a+b=b

Let's say that a=b, well than a-b= 0. Well, how can you factor out a multiplication of 0. You simply can not do that, and so that step is problematic.