Question

prove that 0/0 is 1​

Answer 1

Step-by-step explanation:

0 upon 0 is 1 then one time 0 is so it is correct if 0 upon 0 is 0 then 0 times 0 is 0, so it is also correct if 0upon 0 is undefined than you can't multiply back the first two can not be proved false using this method not can the later since it is not exactly defined as division anyway

Answer 2

Guess, I am gonna prove it for u. ¬‿¬

Provided that you have studied std 8 algebraic equations.

Let's take a random negative number:

$\sf -2 = -2 \\\\\sf \color{green}{ 1}- \color{red}{3} = \color{green}{4} - \color{red}{6} \\\\\sf \color{green}{ 1^2}- \color{red}{1x3} = \color{green}{2^2} - \color{red}{2x3} \\\\\sf Adding~\frac{9}{4}~ on~both~sides \\\\\sf \color{green}{ 1^2}- \color{red}{(1x3)} + \color{purple}{\frac{9}{4}}= \color{green}{2^2} - \color{red}{(2x3)} + \color{purple}{\frac{9}{4}}\\\\\sf \color{green}{ 1^2}- \color{red}{(2x1x \frac{3} {2})} + \color{purple}{\frac{3^2}{2^2}}= \color{green}{2^2} - \color{red}{(2x2x \frac{3} {2})} + \color{purple}{\frac{3^2}{2^2}}\\\\\sf Using~(a^2+b^2-2ab)=(a-b)^2 \\\\\sf We~have: \\\\\sf \bigg(\color{green}1- \color{purple}{\frac{3}{2} \bigg)^2 }=\bigg(\color{green}2- \color{purple}\frac{3}{2} \bigg)^2 \\\\\sf\color{green} 1- \color{purple}\frac{3}{2} = \color{green}2- \color{purple}\frac{3}{2} \\\\\sf \color{green}1- \color{purple}\cancel{\frac{3}{2}} = \color{green}2-\color{purple} \cancel{\frac{3}{2}} \\\\\sf \color{green}1=\color{green}2 \\\\\sf subtracting~1~on~both~sides \\\\\sf 1-1=2-1 \\\\\sf 0=1 \\\\\sf Hence 0/0 =1$

Or

$\sf Let~m=n \\\\\sf Multiplying~ both~ sides~ by ~m ~gives : \\\\\sf m^2=mn \\\\\sf Now, subtracting ~both ~sides~ by ~n^2~ gives: \\\\\sf m^2- n^2= mn-n^2 \\\\\sf Using~algebraic~formula ~(a^2-b^2)=(a-b)(a+b) \\\\\sf (m-n) (m+n) = n(m-n) \\\\\sf \cancel{(m-n)} (m+n) = n \cancel{(m-n)} \\\\\sf (m+n) =n \\\\\sf Now, ~see~ above, ~we ~have~ assumed ~ m=n. \\\\\sf So, (m+n) =n~can~be~written~as~ (n+n)=n \\ \\ \sf ~2n=n~=> 2=1 \\\\\sf Subtracting~1~on~both~side: \\\\\sf 2-1=1-1 => 1=0 \\\\\sf Hence, 1 = 0/0$