@fresh_42 This totally is the case. I agree with you and Feynman. People here don’t explain in simple language and pour lots of information as a reply. This makes hard to understand and then they say we have pointed this or that out many times but you ignored. I don’t ignore. It gets lost into...
I saw.
and this proves my point. “Vast majority” always do what’s convenient. They would not understand inverse of a. Same thing with percentage. People understand 100. They are able to Compare things with 100. That’s why we use percent today.
That’s a mistake or typo.
Isn’t everything just addition?
Subtraction is addition of opposite. Multiplication is repeated addition and division is just opposite of multiplication which is also just addition.
Just because division is used by companies with big names doesn’t qualify division as a concept. It’s still multiplication.
They must have used it because its convenient.
Division has made life easier.
It’s for convenience that we teach division to kids. Actually what we are doing is inverted multiplication.
That’s what he meant by “to give all kids same number of cookie”. Division is for convenience.
If we are literally breaking down things into very basics then we are actually doing repeated addition. ##13## added ##13## times =##169##.
But we can save time and do direct multiplication.
For example what is ##\frac {169}{13} = ?##
This says “When ##169## is divided into ##13## groups how many there are in each group?”
This can be converted into a multiplication problem like this “##13## groups of how many in each group makes ##169##?”
This is ##13 * ? = 169##. It can be solved...
Ok. We define things and combine them to make fundamental statements which can be proved by the definition itself like here.
But I am not satisfied that distance definition doesn’t work to prove axiom.
Translation: Axioms are not that difficult to understand.
But I don’t know why this one doesn’t feel obvious.
By ‘it’ you mean property of absolute value inequality just to be clear.
Why are you guys only using the definition of ##|x|## ? Why can’t we prove the property ##|x|<c## is equivalent to ##-c<x<c## by considering the fact that ##|x|## is distance between ##0## and ##x## or as I am speculating it as distance between ##0## and ##-x## ?
It seems like you are saying...
Hello! Good to see you.
I don’t know why but I have a desire to write ##|x|## as ##x## which is completely wrong since ##|-5| \neq -5##. If we are not careful it becomes ##x<c##. On a number line ##x## would be left of ##c## up to ##-\infty## , if ##c>0##. This doesn’t match with the book. This...
If ##x>0## then ##|x|<c## means ##x<c##
If ##x<0## then ##|x|<c## means ##x>-c##
So for all ##x## , ##|x|<c## means ##-c<x<c## (##x## lies between ##-c## and ##c##).
##|x|## means distance between x and 0. But I am thinking we can also write ##|x|=|0-(-x)|## which means it is the distance between 0 and -x. Am I right?
Ok. 13 can be written in infinite ways. ##|0-13|## , ##|0-(-13)|## , ##|5-(-8)|## , ##|-2-11|##. Similarly if ##x## is a variable in place of a number then ##|x|## is the distance between 0 and x. But how do we add ##-x## to the story to prove ##-c<x<c##?
First lets focus on ##|x|## which is defined as distance between ##x##and ##0##. But if we look into it closely
$$13=|-11-2|$$ which is distance between -11 and 2 but $$13=|11-(-2)|$$ which means this is distance between 11 and -2. Which is it?
In the same way $$x=|x-0|$$ is distance between 0...
But I asked one question regarding books.
That’s a relief.
I prepare myself that now I’ll complete this or that book and will pursue my journey to learn science. I start the book and then I leave because the math is above my level. So I go buying new book. And then this cycle repeats. I start...
Actually I want to revise algebra 1 and 2 and move forward. Learn new maths. But I don’t want to buy new book. I was thinking if this book will do the job for a while?
(I am tired of buying books)
Is it possible to find the solution of ##2x-1=-\sqrt {2- x}## in the start without squaring?
I mean we know the solution of ##x=-1## is ##x=-1## at the start. So it would be good if we declare the solution of the original equation in the beginning if we are doing analogies.
By the way in post #9 I factored eqn ##1## (##(2x-1)^2=2- x##)by splitting the middle term and eqn ##2## (##x^2=1##) by the factoring formula ##A^2-B^2##. I don't think that will make any difference how we factor.
Actually I was searching for you, person whom I can understand because they are talking in the same language as my current maths book. So it becomes very easy to follow up. I don't understand much when people talk in high level language. You know what that means. HaHA :)
Let me give it a try. Lets take ##x=-1##. I added -ve sign to make it analogous to our original equation ##2x-1=-\sqrt {2- x}##
##2x-1=-\sqrt {2- x}## equivalent to ##x=-1=-\sqrt 1##
(Here ##2x-1## expression is ##x## and ##2-x## expression is ##1##)
Solution: x=-1/4 and...
Does this mean for x=1 , ##2(1)-1= -\sqrt{2-1}## is false. x=1 is not a solution.
But as we square the above equation , ##(2(1)-1)^2=(-\sqrt{2-1})^2## , false equation becomes true. So now x=1 is solution to the new equation ?
(Here is the paragraph attached) from book James Stewart.
I am given an equation to solve and I am unable to understand why definition of root doesn’t work both ways. Namely if ##\sqrt 5 = x## means ##x^2=5## then ##x^2=5## doesn’t means ##\sqrt 5=x## It has to do with root and square. Square having unique value but root gives two values. But I don’t...
I do fear that if I don’t go in linear fashion I’ll get stuck in future. I have precalculus book in paperback. I will continue with it for now. Thanks!
That I missed. Good point. And now I understand that ##(\sqrt{2x+1})^2= ({2x+1})^{\frac 12(2)}= 2x+1##.
In this part ( ##x^2=5##) why can’t I just use the definition of radicals ?
In order to write next step in all four equations above l used the definition of radicals. ##\sqrt a=b## means ##b^2=a##. Squaring both sides also works. I don’t know if it’s right. I mean I read that ##(\sqrt a)^2=a##. But I don’t know if we can apply this on expressions.
Main problem is if we...
I am currently learning some maths from “Precalculus by James Stewart”.
I was wondering if that’s ok? Is it ok to just dive straight into it or go back and brush up my algebra 2 ?
I was wondering what are some good textbooks on algebra 2 by the way?
Thank you.
(This is all for the love of physics).
TL;DR Summary: ##(1+ \frac1x)^2 - (1-\frac1x)^2##
##(z+2)^2 -5(z+2)##
Upon simplifying the first I get ##\frac4x##. So isn’t the first expression fractional?
Upon simplifying the second I get a Quadratic expression.