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I don't think [math]1^2 = 1[/math];...
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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?
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>>7852069
Fuck off Howard.
>>
1^2 =/= 2 because sqrt(2)^2 = 2
>>
X^0 +1 = Prime number

All natural numbers are prime
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>>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 ?
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>>7852112

[math]2^2 = 4[/math] so 4 is not prime
>>
>>7852069
Your dick doesn't have an effect either.
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>>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]
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>>7852069
I don't think 4+0=4, I think 4+0=5.
If 4+0=4 then +0 has no effect.
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>>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
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>>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]
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