Beginner math help

>>8229999
>I feel like III is most definitely wrong.
Why? It's correct, which you can
verify by substitution of your results
into the original equation.

Maybe the property you're unsure of is that there's no zero dividers for real numbers. This translates to the fact that if for all real a:
a*b=0
then b=0. Consider fixing x in your expression. The edge case is y=0, in which case x^2=0 or x=0 and so x=0=y=-y satisfies the claim. Now, safely assume y is not zero. Then x-y != x+y. You have two possibilities for your expression to hold, either x-y = 0 or x+y=0. Then it must be that x=y or x=-y for all values.

Nice quads btw>>8229999

1) is wrong because you don't know a priori that x is invertible, but since a is by assumption you can multiply rhs and lhs by a^{-1} to get the result.

2 and 3 are fine, maybe you want to add written explanations of why you are allowed to do each step?

>>8229999
>I.) if ax = a and a != 0, prove x = 1
>a(x^(-1)x) = 1a = a
>x = 1
What?

>>8230024
It actually defines multiplicative inverses in the previous section.

everythings probably fine although I have no idea what you are trying to say with the first answer. just multiply both sides by the multiplicative inverse of a instead and use the distributive property. also for the last one you should show by contradiction that if the product of two numbers is 0, at least one of the numbers must be 0, you gloss over that for some reason.

This site is awesome, simple rules and guides for any type of maths.

http://www.krysstal.com/algebra.html

>>8229999
http://www.krysstal.com/algebra.html

>>8229999
>if ax = a and a != 0, prove x = 1

Given that it is calculus it is assumed that this is the field (R,+,*)

(R,+,*) can be a field if and only if (R-{0},*) is a group.

(R-{0},*) can be a group if and only if there exists a 'neutral' element, lets call it x.

x can only be the neutral element if and only if for all a in the set R-{0}, ax=a

It is also trivial that x is unique, because is the neutral element exists then it must be unique.

Lets consider the set {1} and construct the group (and subgroup of the previous group) ({1},*)

This is a group because 1*1=1 (identity element exists... and also 1 has its symmetric element)

1(1*1) = (1*1)*1 = 1 so it is associative and therefore this is a group.

Given that * is the same operation as before and {1} is contained in the set of all the real numbers, this is a subgroup of (R-{0},*)

The identity element of a subgroup is the identity element of its supergroup

Therefore the identity element of (R-{0},*) is also 1.

Therefore the equation ax=a has the unique solution x=1

QED

>>8229999
Is this for an abstract algebra course? You wouldn't be proving this in a calculus class, would you? I am going to assume that you are working over a field, though the proofs would be the same in a non-commutative division ring. I.) Multiply both sides by the inverse of [math]a[/math] to get [math]x = a^{-1}ax = a^{-1}a = 1[/math]. II.) Use the distributive property. III.) Since [math]x^2 - y^2 = (x - y)(x + y) = 0[/math], you can conclude that [math]x + y[/math] or [math]x - y[/math] is zero. Otherwise, both belong to the multiplicative group, meaning that their product is not zero. Adding the additive inverse of [math]\pm{y}[/math] to both sides completes the proof.

you failures are so exhuberant to answer one question correctly you dont even read the thread before you post your redundant response

>>8230220
It takes time to write these types of responses. Someone might post an answer while you are writing. If you don't set it to auto-update, you might be unaware that another solution was already posted.

>>8229999
Take your pedophile cartoons back to >>>/a/.

>>8230083
Yes, which exist for non-zero elements. Are you sure x is non-zero?

>>8231679
You're only interested in the case where x = 1, since obviously if x = 0 then a*x = 0 by definition. You're proofs should be good OP. Since you appear to be in calculus, you could probably be even less strenuous with the first one. Since a is nonzero you can simply divide both sides by a.There are a myriad of other reasons, namely that for the real numbers 1 is the identity element whose primary characteristic is that a*1 = a for all a in the real number line.

ax = a
ax/a = a/a
x = 1

II) just foil the right side and combine like terms

III) square root both sides and you get +-y
too easy OP