1

I'm currently reading a Huffington Post article by Ed Mazza ("A Glint of Light and a Hint of Life: Mars Is Getting Very Interesting Right Now," June 24, 2019) and I came to a paragraph.

The paragraph says:

It doesn’t appear on any of the images snapped before or after, taken about 13 seconds apart, so if it was an object of some kind it moved quickly. More likely, however, it was nothing too out of the ordinary.

I'm unable to grasp the meaning of the bold part. What could this line mean? Could that mean "it was nothing, which like other stuff out of the ordinary?" I don't get the meaning. Could someone please explain to me what this line means within the above context?

Improve this question

1 Answer 1

Reset to default
1

"Nothing too out of the ordinary" means that the object was probably nothing not-ordinary/abnormal. The object was likely ordinary/normal.

Some synonyms for "out of the ordinary" would include unusual, strange, or unheard of. You can see thesaurus.com for a full list "out of the ordinary" synonyms:

Here is the Merriam-Webster definition:

out of the ordinary (idiom) unusual, different, or strange : not what is considered to be normal
What happened is nothing out of the ordinary.

Improve this answer
5
  • Thanks! In my almost 25 years English related life, I have heard for the first time "out of the ordinary" would be the potential synonym of "ordinary"! haha.
    – user17814
    Jun 24, 2019 at 20:41
  • 1
    "Out of the ordinary" is actually an antonym of "ordinary." Sorry I didn't make that clear!
    – jmrpink
    Jun 24, 2019 at 20:43
  • 1
    Thanks. But the synonym list you gave me helped a lot!
    – user17814
    Jun 24, 2019 at 20:45
  • 2
    It's worth noting that "out of the ordinary" is an established idiom. (You wouldn't hear a native speaker saying "out of the normal" or "out of the usual".)
    – J.R. ♦
    Jun 24, 2019 at 21:01
  • @J.R Thanks for the additional info. Let us forget about the difference of our opinions prior.
    – user17814
    Jun 24, 2019 at 22:41

You must log in to answer this question.