brainlet Spivak help lmoa!

1 cannot be less than 0 since it is a square. Assume 1 is less than 0. Multiplying it by itself will yield a positive number as per the general rule, which is also 1. See the problem

>>8683808
i swear to god im not memeing and im just retarded but
>1 cannot be less than 0 since it is a square.
can you elaborate? why can't we say the same thing about -1? im also confused why he's bringing it up at all

i thought the fact that it doesn't have a negative sign infront of it, and it's not 0, just implies it's positive off the bat

>>8683821
>>8683827
oops i meant to reply to you there

>>8683827
1 is 1^2 so 1 is a square and it has to be positive.
-1 is not a square.

>>8683874
again im not seeing the connection here, why can't i say
>1 is (-1)^2 so -1 is a square and it has to be positive
isn't a square just when you multiple a number by itself? also if what you're saying makes sense, then what does the rule he just mentioned have to do with it?

>>8683887
a^2 > 0 if a is not equal to 0
1 is not equal to 0(see axiom above)
1=1 * 1 thus 1 > 0
[1 = (-1) * (-1) thus 1 > 0 I'm not sure whether he proved (-a)(-b)=ab here]

>>8683890
oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

wow im braindead

because you get 1 as a result, it has to be positive

jesus christ im not jason borne

>>8683894
btw you should forget what addition and multiplication mean.
1 * 1 = 1 by identity law

>>8683897
what did he mean by this
i dont get how those two sentences are related

[math]
(a > 0 \land b > 0) \Rightarrow (a \cdot b > 0) \\
(a < 0 \land b < 0) \Rightarrow (a \cdot b > 0) \\
(a \neq 0) \iff (a^{2} > 0)
[/math]

>>8683897
pls elaborate if ur out there

if a real number different than 0 can be represented as a square of another real number, then it's bigger than 0.
25 can be represented as [math] 5^{2} and (-5)^{2}[/math] therefore 25 > 0;
2 can be represented as [math]\sqrt{2}^{2} and (- \sqrt{2})^{2} [/math] therefore 2 > 0;
1 can be represented as [math] 1^{2} and (-1)^{2} [/math] therefore 1 > 0;
-1 cannot be represented as the square of any real number, therefore it's not bigger than 0.

>>8683964
I mean every statements you write down should based on axiom

>>8683974
you mean not to take addition and multiplication for granted, rather than forgetting what they mean? i dont understand otherwise

File: Apostol.png (30KB, 1355x118px) Image search: [Google]

30KB, 1355x118px

>>8683976

>>8683983
im much too brainlet for this i dont understand at all why i need to forget what addition and mulitplication mean or even what forgetting what addition and multiplcation mean means

>>8683983
wat axioms
i know i said im too brainlet but this is gunna bug me for about the next decade

>>8683808
these retards went about a really complicated way of saying 1>0 when they could've just taken that as an axiom from the start.

>>8684249
Not really because thats not in the field axioms