I''ve dealt with two local jewelers. I told them I was willing to spend up to 7k on a diamond. I figured the ring would cost 1k additional (not sure if that is true). The ring will be a solitaire with either white gold or platinum.
Of course, I want a beautiful diamond.
I''m looking for a one carat (can be a little more or less), either princess or asscher cut (the jury of one is still out), colorless, as low as S1 for princess and as low as VS2 for asscher, and the best possible cut.
Both jewelers are telling me about, and showing me, diamonds in my price range. However, as I learn more, it seems that I can spend quite a bit less and still get a beautiful diamond.
It seems that in order to bring the price up to the 7k, they need to increase the clarity and the size.
Perhaps I shouldn''t have told them my price range. Had I known my specs, I would have asked them to find a gorgeous diamond that meet my specs.
I initially said 7k because I was, for some reason or another, under the impression that that''s what I would need to spend to get a beautiful 1 carat diamond.
Can I get a beautiful diamond for far less than 7k?
Thanks,
Mark
Of course, I want a beautiful diamond.
I''m looking for a one carat (can be a little more or less), either princess or asscher cut (the jury of one is still out), colorless, as low as S1 for princess and as low as VS2 for asscher, and the best possible cut.
Both jewelers are telling me about, and showing me, diamonds in my price range. However, as I learn more, it seems that I can spend quite a bit less and still get a beautiful diamond.
It seems that in order to bring the price up to the 7k, they need to increase the clarity and the size.
Perhaps I shouldn''t have told them my price range. Had I known my specs, I would have asked them to find a gorgeous diamond that meet my specs.
I initially said 7k because I was, for some reason or another, under the impression that that''s what I would need to spend to get a beautiful 1 carat diamond.
Can I get a beautiful diamond for far less than 7k?
Thanks,
Mark