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.

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1^2 =/= 2 because sqrt(2)^2 = 2

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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]

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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

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>>7852069

Your dick doesn't have an effect either.

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>>7852069

>terrence howard meme

>Salvia officinalis

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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.

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>>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.

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>>7852112

2^0 +1 =2 which is prime

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>>7853293

But 0 is just nothing, so of course adding nothing has no effect.

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>>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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