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I actually believe that it can be proven.

Proof: Let $R$ be an element of $SO(n)$. So, in 3D this is just the usual rotation operator. We start with a definition

Definition: We call $q$ a scalar under a rotation if and only if it transforms under the trivial representation of the rotation group. That is, if $q'=Rq=q$.

Now, suppose that $q\in \mathbb{R}$. Then, it is a rank zero tensor. We may write $q = q e^1_ie^1_j \delta^{ij} $. It can be shown (I leave it to you) that under rotations that

$$q= q e^1_ie^1_j \delta^{ij} \to q e^1_{i'}e^1_{j'} \delta^{i'j'} = q e^1_ie^1_j \delta^{ij} = q. $$

Hence, we have used the fact that the kronecker delta transforms as a scalar under rotations to show that if $q$ is a real number then it transforms as a scalar under rotations.

I actually believe that it can be proven.

Proof: Let $R$ be an element of $SO(n)$. So, in 3D this is just the usual rotation operator. We start with a definition

Definition: We call $q$ a scalar under a rotation if and only if it transforms under the trivial representation of the rotation group. That is, if $q'=Rq=q$.

Now, suppose that $q\in \mathbb{R}$. Then, we may write $q = q e^1_ie^1_j \delta^{ij} $. It can be shown (I leave it to you) that under rotations that

$$q= q e^1_ie^1_j \delta^{ij} \to q e^1_{i'}e^1_{j'} \delta^{i'j'} = q e^1_ie^1_j \delta^{ij} = q. $$

Hence, we have used the fact that the kronecker delta transforms as a scalar under rotations to show that if $q$ is a real number then it transforms as a scalar under rotations.

I actually believe that it can be proven.

Proof: Without loss of generality, we just prove it for the case of rotations in 3D. We start with a definition

Definition: We call $q$ a scalar under a rotation if and only if it transforms under the trivial representation of the rotation group. That is, if $q'=Rq=q$.

Now, suppose that $q\in \mathbb{R}$. Then, it is a rank zero tensor. We may write $q = q e^1_ie^1_j \delta^{ij} $. It can be shown (I leave it to you) that under rotations that

$$q= q e^1_ie^1_j \delta^{ij} \to q e^1_{i'}e^1_{j'} \delta^{i'j'} = q e^1_ie^1_j \delta^{ij} = q $$

Hence, we have used the fact that the kronecker delta transforms as a scalar under rotations to show that if $q$ is a real number then it transforms as a scalar under rotations.

I actually believe that it can be proven.

Proof: Let $R$ be an element of $SO(n)$. So, in 3D this is just the usual rotation operator. We start with a definition

Definition: We call $q$ a scalar under a rotation if and only if it transforms under the trivial representation of the rotation group. That is, if $q'=Rq=q$.

Now, suppose that $q\in \mathbb{R}$. Then, it is a rank zero tensor. We may write $q = q e^1_ie^1_j \delta^{ij} $. It can be shown (I leave it to you) that under rotations that

$$q= q e^1_ie^1_j \delta^{ij} \to q e^1_{i'}e^1_{j'} \delta^{i'j'} = q e^1_ie^1_j \delta^{ij} = q. $$

Hence, we have used the fact that the kronecker delta transforms as a scalar under rotations to show that if $q$ is a real number then it transforms as a scalar under rotations.

I actually believe that it can be proven.

Proof: Without loss of generality, we just prove it for the case of rotations in 3D. We start with a definition

Definition: We call $q$ a scalar under a rotation if and only if it transforms under the trivial representation of the rotation group. That is, if $q'=Rq=q$.

Now, suppose that $q\in \mathbb{R}$. Then, it is a rank zero tensor. We may write $q = q e^1_ie^1_j \delta^{ij} $. It can be shown (I leave it to you) that under rotations that

$$q= q e^1_ie^1_j \delta^{ij} \to q e^1_{i'}e^1_{j'} \delta^{i'j'} = q e^1_ie^1_j \delta^{ij} = q $$

Hence, we have used the fact that the kronecker delta transforms trivially under rotations to show that if $q\in \mathbb{R}$ is a real number then it transforms as a scalar under rotations.

I actually believe that it can be proven.

Proof: Without loss of generality, we just prove it for the case of rotations in 3D. We start with a definition

Definition: We call $q$ a scalar under a rotation if and only if it transforms under the trivial representation of the rotation group. That is, if $q'=Rq=q$.

Now, suppose that $q\in \mathbb{R}$. Then, it is a rank zero tensor. We may write $q = q e^1_ie^1_j \delta^{ij} $. It can be shown (I leave it to you) that under rotations that

$$q= q e^1_ie^1_j \delta^{ij} \to q e^1_{i'}e^1_{j'} \delta^{i'j'} = q e^1_ie^1_j \delta^{ij} = q $$

Hence, we have used the fact that the kronecker delta transforms as a scalar under rotations to show that if $q$ is a real number then it transforms as a scalar under rotations.