Prove that it is impossible for a number greater than $7$ to appear on the blackboard.

Question

The numbers $2,3$ and $6$ are initially written on a black board and you are allowed to replace two of them say $x$ and $y$ with the two numbers $\frac{3}{5}x + \frac{4}{5}y$ and $\frac{4}{5}x - \frac{3}{5}y$. Prove that it is impossible for a number greater than $7$ to appear on the blackboard.

I took a linear algebraic approach and noted that this operation is equivalent to the linear transformation $$\begin{pmatrix}\frac{3}{5}&\frac{4}{5}\\ \frac{4}{5}&-\frac{3}{5}\end{pmatrix}$$ acting on $\begin{pmatrix}x\\ y\end{pmatrix}$.

Now this is pretty much a rotation and a scaling and so for a number greater than $7$ to appear on the board one of the components of the scaled vectors should be $7$, but if we are starting with numbers less than $7$ this is impossible as we are only getting smaller values?

Is this correct? I guess we also cannot get $7$ ever either.

Answer 1

I like your linear algebraic approach, but when you write that this is ``pretty much a rotation and a scaling'', what are you scaling by?

In fact, this is just a rotation.

If we think of $\begin{pmatrix} x \\ y \end{pmatrix}$ as a point in $\mathbb{R}^2$, then it is a distance of $\sqrt{x^2+y^2}$ away from the origin. On the other hand, the point $\begin{pmatrix} \frac{3}{5}x+\frac{4}{5}y \\ \frac{4}{5}x-\frac{3}{5}y \end{pmatrix}$ is a distance of

\begin{align} \sqrt{\left(\frac{3}{5}x+\frac{4}{5}y\right)^2 + \left(\frac{4}{5}x-\frac{3}{5}y)^2\right)^2} &= \sqrt{\frac{9}{25}x^2 + \frac{12}{25}xy + \frac{16}{25}y^2 + \frac{16}{25}x^2 - \frac{12}{25}xy + \frac{9}{25}y^2 } \\ &= \sqrt{ \frac{9+16}{25}x^2 + \frac{9+16}{25}y^2 } \\ &= \sqrt{ x^2 + y^2 }. \end{align}

Any point $(x,y)$ such that either $|x|$ or $|y|$ is greater than $7$ will necessarily be a distance of at least 7 away from the origin; for example, if $x\geq 7$, then $\sqrt{x^2+y^2} \geq \sqrt{49+y^2} \geq \sqrt{49} = 7$.

However, using $2,3$ and $6$, the furthest point from the origin that we can create is $(3,6)$, which is only $\sqrt{45}$ away from the origin!

Answer 2

If to understand the problem that we can replace any two of numbers with new ones with the presented formula and to leave unchanged a third one we in fact acting in 3d space on vector $$v=\begin{pmatrix} 2 \\ 3 \\ 6 \end{pmatrix}$$ with the use of 3-d dimensional orthogonal transformations with fixing the selected $x,y,z$ coordinate of the vector in a single step,

for example orthogonal transformation with stable $z$ coordinate is represented by:

$$Q_z=\begin{pmatrix}\frac{3}{5}&\frac{4}{5}& 0\\ \frac{4}{5}&-\frac{3}{5}& 0\\ 0&0& 1\end{pmatrix}$$

(here it's not a proper 3-d rotation because determinant of the matrix above is equal to $-1$)

Matrices of other possible orthogonal transformations:

$$Q_x=\begin{pmatrix} 1& 0& 0\\ 0&\frac{3}{5}&\frac{4}{5} \\ 0 & \frac{4}{5}&-\frac{3}{5} \end{pmatrix}$$

$$Q_y=\begin{pmatrix}\frac{3}{5}& 0 & \frac{4}{5} \\ 0&1& 0\ \\ \frac{4}{5}& 0 &-\frac{3}{5} \end{pmatrix}$$

I understand we can change type of transformation in every step.

Any composition of orthogonal transformations acting on $v$ vector is a some orthogonal transformation.

$$w=Q_n\dots Q_2 Q_1 v$$

Orthogonal transformation preserves length of a vector what is in this case $ \Vert v \Vert =\sqrt{2^2+3^2+6^2}=7=\Vert w \Vert $.

The desired conclusion follows from it.