Αποτελέσματα Αναζήτησης
27 Αυγ 2015 · $$ \frac{0}{0}=\frac{2\cdot 0}{1\cdot 0}=\frac{2}{1}$$ The problem is in the first line, when you write $\frac{0}{0}$, which is undefined. Of course you could define it, but then it would be equal to every fraction since $$\frac{a}{b}=\frac{c}{d}$$ if $ad=bc$, and if $a=b=0$, then this is always true, since for any $c$ and $d$, $0\cdot d=0\cdot c$.
- Prove that zero multiplied by zero is equal to zero.
The same way you could prove that $r\cdot0=0$, for each $r$....
- Prove that zero multiplied by zero is equal to zero.
1 Δεκ 2015 · Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number. $\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$ $ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$ Q.E.D.
5 Ιαν 2024 · In this short video, Mr Ong shows how to prove that 0/0 = 2. Can you find the misconception? Let us know.
23 Φεβ 2016 · The same way you could prove that $r\cdot0=0$, for each $r$. Indeed, let $r\cdot0=q$. Then $q=r\cdot0=r\cdot(0+0)=r\cdot0+r\cdot0=q+q$. Hence $q=q+q$, thus $0=q-q=q+q-q=q$. To make the proof look like yours, start with $0\cdot0$ and transform this to $0$. Indeed $0\cdot0=(a-a)(a-a)=a^2-a^2+a^2-a^2=0+0=0$.
17 Νοε 2024 · In todays video, I will be teaching you how to solve an interesting problem.
11 Νοε 2017 · In practice, division by #0# is (almost) always undefined and #0/0# is an indeterminate form. The true version of the premise assumed above might be "Any non-zero number divided by itself is #1# ". Answer link
18 Σεπ 2016 · While division by 0 is undefined in most cases, in this specific scenario of 0/0 = 2, we are not dividing by 0 itself. Instead, we are approaching 0 from both the numerator and denominator, which results in a limit of 2.
