Let
Proposition
There is only one vector in
satisfying the conditions of the vector. In quantifiers,
Step-by-step proof
What is the condition that we seek to show? Uniqueness, that is, that there is only one element that satisfies the given property: when added to any vector
, the sum result should be . Can we check the condition directly? This would require us to take all vectors in
and check that only one satisfies the condition. Instead, we could check that the opposite statement does not hold, that is, reason by contradiction. What does the opposite statement look like? Formally speaking, the opposite statement to uniqueness is the following: either there are no vectors with the property, or there are more than one. However, we already know that the requirements for
to be a vector space guarantee us at least one element satisfying the vector condition. Therefore, in our case the opposite statement would be that there are more than one vectors satisfying the condition that, when added to any vector , the sum result is still . If there are more than one, then there is at least two different ones! Let us denote them and . Our proof should use the property that each of them satisfies the property. How to do that? Let us consider the following sum:
. From one side, , because the vector satisfies the property. At the same time, because the vector also satisfies the property, which gives us . This contradicts the assumption that these vectors are different from each other.
Proposition
For every vector
there is a unique vector satisfying the property of being an additive inverse. In other words, the choice of is unique. In quantifiers,
Proof
Similarly to the previous proof, let us proceed using reasoning by contradiction. We assume that there are two different vectors
and that satisfy the condition to be inverses. Then The resulting equality contradicts that
and are different.
Proposition
Proposition
Proposition
Proposition