I don't think [math]1^2 = 1[/math]; I think [math]1^2 = 2[/math].
If [math]1^2 = 1[/math] then [math]1 \cdot [/math] has no effect.
How can rationals be countable if real numbers are uncountable?
>>7852069
Fuck off Howard.
1^2 =/= 2 because sqrt(2)^2 = 2
X^0 +1 = Prime number
All natural numbers are prime
>>7852103
[math]1^2 \neq 1[/math] because [math](-1)^2 = 1[/math]
What should [math] 1 \cdot [/math] be then ? The number of offspring when you fuck a goat ?
>>7852112
[math]2^2 = 4[/math] so 4 is not prime
>>7852069
Your dick doesn't have an effect either.
>>7852069
>terrence howard meme
>Salvia officinalis
x^0= 1
1^2 = (x^0)^2
1^2= x^(2*0)
1^2 = x^0
1^2 = 1
I'm just an engineer student tho, not very good with proofs, so this may just be bullshit.
>>7852103
>1^2 =/= 2 because sqrt(2)^2 = 2
So [math]\sqrt{2} = 1[/math]. You wouldn't say that [math]2^2 \neq 4[/math] because [math](\sqrt{4})^2 = 4[/math]
>>7852069
I don't think 4+0=4, I think 4+0=5.
If 4+0=4 then +0 has no effect.
>>7852114
Sure, but:
[eqn]
\sqrt{1} = \pm 1
[/eqn]
So your argument is invalid.
>>7852112
2^0 +1 =2 which is prime
>>7853293
But 0 is just nothing, so of course adding nothing has no effect.
>>7853272
(4^(1/2))^2 = 4^(2/2) = 4
Yes you can, tard
>>7853598
[math]\sqrt{2}^2 = 2[/math]
[math]\sqrt{4}^2 = 4[/math]
If [math]2^2 = 4[/math] then [math]\sqrt{4} = 2[/math].
IF [math]1^2 = 2[/math] then [math]\sqrt{2} = 1[/math]