A Eulerian circuit is a loop, but trees have no loops. QED.
(Although that is just a shorthand for what you wrote. I can't think of an easier way...)
If G is a tree it has no cycles
Assume G is a tree
If G has an Eulerian Circuit then the Eulerian Circuit Must repeat a vertex other than the initial vertex for G to remain a Tree
therefore there is one vertex which has no bridge.
-> it is not a tree
This is not explained as well as I could do.
Maybe there is a much simpler way of showing this.
Please advise thanks you
I have found another solution which doesnt rely on circuits or cycles
The number of edges in a tree is n-1
The sum of the degress of the vertices is then 2(n-1) = 2n-2
To have a eulerian graph The degree of every vertex must be even
So Sum of the degress of vertives >= 2n
But 2n-2 is never >= 2n
therefore it cannot be a tree.