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I'm writing an essay and we can't use "to be" verbs. Is "become" a to be verb? Thanks.

2007-05-26 02:27:28 · 4 answers · asked by Martha Deacon 1 in Society & Culture Languages

Can I use "may be"?

2007-05-26 03:13:05 · update #1

4 answers

The eng verb to be has eight forms the main three are 'am, is, are' the other five are:

'bin, been, was, were, and be[ we might add being(in continuous tenses, or in passives, that is)'

'become', however; is a verb to do, which is another category of verbs.

'may', like:' will, might, shall, etc...', are modal auxiliary verbs ,i.e. they help the main verb in a sentence or they can represent as a main verb themselves. [but i think you know that!]

Anyway, according to what i've mentioned above, i don't think you should use 'maybe' as an alternative to verb to be, because it'll already be one even if used with another verb.

Good luck on your essay!

2007-05-26 04:21:29 · answer #1 · answered by Judy 5 · 0 0

if your teacher means "to be verbs" to be anything more that "am is are was were," then become is certainly included.

one possibily as for what "to be verbs" could mean:

the verbs that can be swaped by "to be" in sentences like:

she is angry
she is a general
he is a man

in all these you can say becomes instead of is, thus classing become a to be verb.

another possibility is that "to be verb" simply means "be been being am is are was were," if that's the case, become is not a "to be verb".

2007-05-26 07:18:49 · answer #2 · answered by Anonymous · 0 0

As far as I know, "to be" is the only "to be" verb. (am/are/were/was/will be/ is being/ has been etc) Did your teacher give an explanation?

No, "may be" is still "be". Would "seem" help?

2007-05-26 02:35:04 · answer #3 · answered by Goddess of Grammar 7 · 0 0

No, it isn't. I think the point of your teacher's assignment is to show how hard it is to write without using some form of "be" in English. It is much easier in such diverse languages as Russian and Arabic.

2007-05-26 02:43:56 · answer #4 · answered by obelix 6 · 0 0

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