For #11: your calculator answer is inappropriate when it answers 58008

I'm trying to actually solve this. I saw the answer for #1. It says that if you consider 1/0 to infinity, 1/0 * 0 = 1, if you cancel 0 or 1/0 * 0 = infinity * 0 = 0. This leads to ambiguity of saying (any number)/0 is infinity.

I think #2 is a bit too sweeping of a statement. It sort of reminds me how Richard Feynman [jokingly] proposed a GUT, which is to round up and state all equations of physics in the form $X=0$ and then multiply them all to get $\prod { X } =0$. Yes, it'd be a GUT, but it wouldn't be very illuminating, wouldn't simplify anything, and wouldn't help us learn anything new.

The equals sign, $=$, is actually used in a variety of context, as for example, in an identity, where both sides are always the same, or in an assignment, where a variable is given a specific value, or in an definition, as for example a Log function is defined to be the inverse of an exponential, etc. Not all equals are equal. One who programs computers would be familiar with these kinds of issues. To say that "solving problems typically requires finding equivalent statements that simplify the problem", suggesting that problem solving is a chain of equal quantities from unsolved to simple solved is just grossly oversimplifying the practice of doing mathematics.

10)

to find the root of

$x^{2} + 1 = 0$

13)

x any be real numbers

$- x . - y + ( -x . y) = -x( - y + y) =0$

thus negative . negative is equal to positive

sir is this right?

12) let us suppose that there are n number of 9

we can write it in this way

$\frac{9999.....}{10^{n}}$

$= \frac{10^{n} - 1}{10^{n}}$

$= 1 - \frac{1}{10^{n}}$

thus as n will go on increase $\frac{1}{10^{n}}$ go on decrease (tends to 0)

thus $= 1$

(1)Proving that you cannot divide by 0: First I will prove that you cannot divide a $\textbf{non-zero}$ number by 0.Assume that proving by 0 is possible.Let $a$ be any non-zero real number.Then $\frac{a}{0}$ results in a number $b$ which can be written as: $\frac{a}{0}=b\\a=b\times 0$ Anything times 0 is 0,so $b$ times 0=0.But then the equation becomes $a=0$.But we have already fixed that a is non-zero.Therefore,we have arrived at a contradiction.Therefore we cannot divide non-zero numbers by 0. Second part of proof:Proving that 0/0 is undefined: Let 0/0 be equal to $c$ Then: $\frac{0}{0}=c\\0=c\times 0$ But LOOK:$c$ can be any number;it can be $0,1,\frac{1}{2},\pi\;etc$ $\textbf{ANY NUMBER}$.But a division can only have 1 answer.So it is proved that $\large{\textbf{YOU CANNOT DIVIDE BY 0}}$

(4)1/60,0.00156,0.0015,0.002,0.001

(5)$\sqrt{2}\times\sqrt{3}$,which cannot be expressed as repeated addition.In general $\sqrt{a}\times\sqrt{b}=\sqrt{ab}$ cannot be expressed as repeated addition if both $a$ and $b$ are not perfect squares. $\textbf{Explanation:}$You can express $a_1\times a_2 \times a_3\times\dotsm a_n$ as repeated addition if any one of the numbers from $a_1$ to $a_n$ is an integer.But this is not necessarily true.

(6)You cannot order 30 tables because 3 people will still be left standing so you have to order 31 tables.

(7)For example,if a teacher assigns a final grade by using the mean of a students with marks:19/20,18/20,20/20,19.5/20 and one really bad 10/20 (Which does happen) the student's final grade will be really low because of that 10/20.In such cases,it is better to use other types of averages.

(8)$F=\frac{Y}{3}$

(10)To find solutions to those equation which had no real solutions.

(12)True $\textbf{Proof:}$Let: $0.999\dots=x\\9.9999\dots=10x\\10x-x=9.999\dots-0.999\dots\\9x=9\\x=\frac{9}{9}=\boxed{1}$ (13)$0=(-1)\times0\\=(-1)\times[1+(-1)]\\=(-1)(1)+(-1)(-1)\\=-1+(-1)(-1)=0\\-1+(-1)(-1)=0+1\\\boxed{(-1)(-1)=1}$ So $(-x)\times(-y)=(-1)\times(-1)\times x\times y=1\times xy=xy$

for number 12: its true using geometric infinite series.

Multiplication is repeated addition only for integers, we cannot define it for real numbers where we cannot express it as such,., in fact if multiplication was really just repeated addition,,

then x^2=x+x+x+x......x (x times for all real no.) and on differentiating, we get 2x=1+1+1+1+... (x times) = x in which case we get 2x=x for all real number which is false

I disagree with #1.

Depends on how you define "dividing," you can actually get a "defined" result.

In the Riemann Sphere,

s

For this set of arithmetic, one may yield, for any nonzero $a$, $\frac{a}{0}=$ complex infinity.

1)

Using binary operation on set of complex numbers , when we divide any complex number by zero it does not belongs to its set

Using binary operation on set of complex numbers , when we multiply any complex number by zero it belongs to its set

therefore we can multiply any number with zero but can't divide

IS this right?

The first statement says that "You can't divide by zero." You can simply say

$\frac{x}{0}$

is impossible because...

$y \times 0 = x$ ,

in which where $x \neq 0$. No such real number can stand for $x$.

help me learn here:

@no.1- zero implies "infinite space where infinite number possible can fit." You can't possibly divide something or anything with an infinite space where infinite number possible could exist.

Am I making sense in my statement?

1) we cannot divide by zero because the result is undefined where as if we multiply by zero then the result itself is zero.

6) No. Of tables required= 31

12). True

12) 0.999.... = 1 It is only a matter of notation. No mystery at all.

1 we cannot divide by 0 but multiply because in division we not only have 1 but 2 conditions. in multiply it creates no problem because the answer is always 0 and we don;t need to do discussion over that , but in division we have to find the nearest multiple of any positive or negative no. for eg. - 6 . but even if we multiply 0 *10000000000 we know we would never reach to 6. and hence it just results to be ∞