
property of equality
state the property which is a reason for each statement.
a. if a<0 and b> 0, then a<b.
b. if x>7, then x is not equal to 7.
c. if 5>9 and 2<3, then 7<12.
d. if m = 10 and n= 6, then m+n = 6
e. if squareroot of a = 5 is true, then square root of a + 4 is false
my answer
a.transitive property
b.trichotomy property
c.subtraction property of equality
d.addition property of equality
e. ?????
iF any could know the exact answer please check my work? thanks....

Are you sure you typed c and d and e correctly? Doublecheck those problem statements, if you would, please. Definitely agree with a and b's answers.

thank you sir for your time looking and checking my work. sir that is exactly the problem ,taken from the book of geometry ( Edwin E. Moise, Floyd L. Downs Jr.Addison Wesley) page 27 problem set number 2. thank you sir for your effort..God bless

Hi jpmath,
You need to correct c d and e as noted by Ackbeet. You are making typing errors
bjh

Reply to jpmath2010:
Well, assuming you're typing the problem statements correctly, here's what I would say:
c. This statement is what we call in logic "vacuously true". The if part is false (it is not true that 5>9), and hence you're allowed to conclude anything whatsoever. Now, if the problem statement is supposed to be this:
If 5<9 and 2<3, then 7<12 (the difference is the first inequality),
then I would say the additive property of inequalities is what works for you there.
d. This is quite simply not true in normal arithmetic. Now, if you're doing modular arithmetic, you might be able to say that. (modulo 10, for example, although then you wouldn't have m = 10, but m = 0). In normal arithmetic, if m = 10 and n =6, then m + n = 16, not 6.
e. This is a nonsensical statement. It simply doesn't parse mathematically. I suppose you could take the square root of an equation (a = 5), but then why would you compare that in a truthfunctional sense with the square of something that's not an equation (a+4)? Did you mean this:
If $\displaystyle \sqrt{a}=5$ is true, then $\displaystyle \sqrt{a}=4$ is false?