# Contradictions $\ldots$$\ldots$$\ldots$

This is a note for discussing contradictions and sharing facts, links opinions, info and more.

• $\infty \times 0 = ?$

• $Is \ \frac{1}{0} = \infty ?$

• $0^{0} = 1, \ but \ why?$

Share more contradicting problems so we can discuss and debate upon their answer!

Note by A Former Brilliant Member
11 months, 1 week ago

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Percy, here's my opinion on the value of $\frac{1}{0}$, and here's my general opinions on working with $\infty$ (my comment, not the discussion).

- 11 months ago

# This is true for all numbers divided by 0, so division by 0 is just undefined......

- 11 months, 1 week ago

If $\frac{1}{0}$ "is" $= ∞,$ then we have an answer for $∞ × ,0$ which will be $\boxed{1}$

- 11 months, 1 week ago

But we know that $0 \times x$ is always $0$ with any value of $x$.

- 11 months, 1 week ago

@Percy Jackson

Only with finite values of $x$

- 7 months, 1 week ago

Yes, that's also true...

- 7 months, 1 week ago

$x$ is a NUMBER and $\infty$ is NOT a NUMBER $\therefore$ $x$ alway have to be finite

- 7 months, 1 week ago

And if you are talking about : $\boxed{\lim_{x\to\infty}(x\times 0)}$ then, $\lim_{x\to\infty}x\times 0 = 0$

- 7 months, 1 week ago

1/0 is undefined because you are basically asking what number times 0 or how many zeroes should you add so that you get 1, even if you add infinite zeroes the answer would be 0.

- 11 months, 1 week ago

If $\lim_{n\to 0}n^n=0\cancel{\Rightarrow}0^0=1$

The result will be different if you include complex numbers (something like $\lim_{a\to0}\lim_{b\to0}(a+bi)^{a+bi}$ or $\lim_{a\to0}(a+ai)^{a+ai}$ or $\lim_{a\to0}(ai)^{a}$, etc ).

$\therefore0^0$ is undefined.

Also if $\lim_{x\to a}f(x)=k\cancel{\Rightarrow}f(a)=k$

- 11 months, 1 week ago

@Hamza Anushath see this if you don't believe me.

- 11 months, 1 week ago

Link Isn't the topic still debatable?

- 11 months, 1 week ago

What about this? Assume $\frac{0}{0}$ is undefined. Then $\frac{1}{0} = \infty$, but $0 \times \infty \neq 1$, since to obtain this, we would have to multiply $\frac{1}{0}$ by $0$, resulting in $\frac{0}{0} \times 1$, which would then be undefined.

- 7 months, 1 week ago

But why $\dfrac{1}{0}$ must be $\infty$?, why not $\red{UNDEFINED}$

- 7 months, 1 week ago

Also $f(x)=\frac{1}{x}$ is discontinuous at $x=0$ so you can't give $f(0)$ any value

- 7 months, 1 week ago

This is true, but only if we assume that there is both a positive and negative value of infinity. What if we have a single point of infinity where everything ends, similar to $0$, where everything begins. Then we would get the situation I described in this discussion. Please note this is all speculation on my part. Definitely fun to think about!

- 7 months, 1 week ago

Percy, 0^0 is not equal to 1...

- 9 months, 4 weeks ago

- 8 months, 2 weeks ago

Well, I just watched a video by Eddie Woo on $0^{0} = ?$ and he solves the question by using limits. He says that it is undefined, but it seems as if its one, so we have agreed upon that value, until we find a better solution. He shows that $\lim_{x \to 0} x^{x} = 1$ by calculating x^x for decimals and showing that is approaches 1.

- 7 months, 1 week ago

$(\lim_{x\to a}f(x)=A)\cancel{\Rightarrow}(f(a)=A)$ $(f(a)=A){\Rightarrow}(\lim_{x\to a}f(x)=A)$

- 7 months, 1 week ago

Also you are looking only for Real limits not Complex limits

- 7 months, 1 week ago

- 7 months ago