7
Well, its a definition in some sense.
One can define the set ${\Bbb N}_0$ of natural numbers by the Peano axioms. They basically say that (1) $0\in{\Bbb N}_0$, (2) $\nu:{\Bbb N}_0\rightarrow {\Bbb N}_0$ is a one to one function with $\nu(x)\ne0$ for all $x\in {\Bbb N}_0$, where $\nu(x)$ is called the successor of $x$, and (3) the induction axiom: If a ...
4
$1+1=2$ is not a fact in general.
Consider the group $\mathbb{Z}_2$ under usual operation.
For this case, you'll get $1+1=0$. The case you are considering is of $\mathbb{Z}$ which is infinite cyclic group and yes here under usual operation $1+1=2$ is true.
This is just a simple example considering group structure there are other examples also. you need to ...
2
One could dispute your colleague's claims. For example, adding two drops of water together still gives you drop of water, which is also observed in nature. The question is quite philosophical and cannot be answered without more information, you could say that it depends on exactly what you mean... You may find the following article on counting from Rafael ...
2
It all depends on what you mean by 1, 2, and +. One can do something like Peano arithmetic and say that we have $0$, a $+1$ operation, and literally define $2=1+1$. This doesn't require any sort of relation to nature, and it makes the statement true regardless of anything else.
But I think the intuition people have about counting comes from counting ...
1
All pairs must be of the form $(10^k, b)$, where $0\leq b < 10^k$.
If $a$ is not a power of $10$, let $M$ be a power of $10$ which is larger than both $a$ and $b$. Consider $X_l = \sum_{i=0}^l M^i$, then $s_{aX_l+b} - s_{X_l}$ is different for different $l$.
If $a$ if a power of $10$ and $b \geq a$, then take $Y_l = 10^l-1$. $s_{aY_l+b} - s_{Y_l}$ is ...
1
This seems pretty straightforward.
$V=D/T\rightarrow T=D/V$
So
$T_2-T_1 = D_2/V_2 - D_1/V_1$
Use $D_2=D_1$, simply solve for $D_1$ given the known time difference and the known velocities.
1
With a reasoning from elementary school:
The difference is $16\,\%$ of that number. So, $1\,\%$ is $16$ times less, i.e. $\frac{448}{16}=28$, and $59\,\%$ is $59$ times the latter:
$$59\times 28=1652.$$
1
As mentioned in the comments, $0.87x - 0.71x = 448$. Then you can use the unitary method:
$$0.16x=448$$
$$0.01x=28$$
$$0.59x = \ ?$$
1
Hint:
$$10a+b=a^2+(a+b)^2\iff2a^2+2a(b-5)+b^2-b=0$$
Discriminant $$=4(b-5)^2-8(b^2-b)=-4b^2-32b+100=4(25-16b-b^2)=4(89-(b+8)^2)$$ has to be perfect sqaure $=(2c)^2$(say)
$$89=c^2+(b+8)^2$$
As $b\ge0,b+8\ge8$
Again, $(b+8)^2=89-c^2\le89\implies b+8\le\sqrt{89}<10$
1
Let $n$ be an arbitrary integer. Divide $n$ one after another by $-2$ and take the remainders (in $\{0,1\}$): $n=q_{0}(-2)+n_{0}$, $q_{0}=q_{1}(-2)+n_{1}$, .... Since $|n|\geq|q_{0}|\geq|q_{1}|\geq\cdots$ and if $|q_{i}|=|q_{i+1}|$ then $q_{i}=0$ or $q_{i}=-1$, we have some N such that $q_{N}=0$. Then $n=\sum_{i=0}^{N}n_{i}(-2)^{i}$.
1
"BODMAS" may well be taught before students have even encountered the notion of exponentiation. In that context it really should be "BDMAS"*, but of course this is not readily pronounceable so an "O" was added. The explanation I was given at school was that this stood for "of" (i.e. the other things can be inside the brackets), which doesn't really mean ...
1
Result: If $p>3$ is a prime number and $x$, $y$ and $z$ are positive integers such that
$$x^3 + y^3 + z^3 - 3xyz = p,$$
then if $p\equiv1\pmod{3}$ we have, after permuting $x$, $y$ and $z$, that
$$(x,y,z)=\left(\tfrac{p-1}{3},\tfrac{p-1}{3},\tfrac{p+2}{3}\right),$$
and if $p\equiv2\pmod{3}$ we have, after permuting $x$, $y$ and $z$, that
$$(x,y,z)=\left(\...
1
As @Hamza Bhati mentioned,
When you multiply two negative numbers, in the last term you have to
write 2's complement of the first number.
I'd like to explain the reason: taking your example, 1011x1101=1011x(1000+0101)=1011x1000+1011x0101, you shall use normal multiplication on the part of 0101, while you shall use 2's complement on the part of ...
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