for all real numbers a and b, if a+b<200, then either a,100 or b<100.?

Suppose the contrary, that it is possible to find a>100 and b>100 so a+b<200.

Then a = 100 + a' , b = 100 + b', where a'>0 and b'>0.

You can write:

a + b = 100 + a' + 100 + b' = 200 + a' + b'

As a'>0 and b'>0 => 200 + a' + b' > 200 = > a+b >200 which is a contradiction with the premise of the problem, so either a<100 or b<100

Take a situation where a=b=100.

a+b will be 200=200 and 200 is not<200.This is a contradiction.

Now take a case where both a>100,b>100.

Here the sum a+b should obviously be greater than 200 as a>100 and b>100(Contradictory).

If you take a<100,b<100 a+b would obviously be less than 200(Contradictory point).

Now take another case where a<100 and b>100.

If b is greater than 200 which is also greater than 100, a can be given a negative value so that sum is less than 200(Negative because real numbers contain negative numbers).

Similarly it holds good for a>100 and b<100.

Therefore we can say that for all real numbers a and b, if a+b<200, then either a,100 or b<100.

When you negate a statement, "or" becomes "and" and "<" becomes ">=" (greater than or equal to)

So negating "either a < 100 or b < 100" gives "both a >= 100 and b > = 100"

So suppose both a >= 100 and b > = 100.

Then a + b >= 100 + 100 = 200. But we're assuming that a + b < 100. We have a contradiction!