Hey can you prove ?
if you can't do it then ask me I can prove it.
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Answered by
1
a-b=b-a
multiplying both sides by zero...
0*(a-b)=0*(b-a)
0=0
proved
multiplying both sides by zero...
0*(a-b)=0*(b-a)
0=0
proved
amaan88:
no without multiply by zero
Answered by
2
it is a little complicated
Remember that "A−BA−B" means "The set of things in AA but not in BB." So "A−B=B−AA−B=B−A" means "The things in AA but not BB are exactly the things in BBbut not AA."
Now, for which AA and BB is this equation true? As always, when you're trying to understand new abstract concepts (in this case, set difference and Boolean operations in general) it's best to try some examples. Does the equation A−B=B−AA−B=B−A hold for A={1,2,3}=BA={1,2,3}=B? What about A={1,2,3},B={1,2}A={1,2,3},B={1,2}? What about A={1,2,3},B={2,3,4}A={1,2,3},B={2,3,4}?
Based on these examples, you should be able to make a good guess at what the answer should be. Now, try to prove it! (As usual, this will look like "Assume x∈A−Bx∈A−B. Then [stuff]. So x∈B−Ax∈B−A. etc.")
Remember that "A−BA−B" means "The set of things in AA but not in BB." So "A−B=B−AA−B=B−A" means "The things in AA but not BB are exactly the things in BBbut not AA."
Now, for which AA and BB is this equation true? As always, when you're trying to understand new abstract concepts (in this case, set difference and Boolean operations in general) it's best to try some examples. Does the equation A−B=B−AA−B=B−A hold for A={1,2,3}=BA={1,2,3}=B? What about A={1,2,3},B={1,2}A={1,2,3},B={1,2}? What about A={1,2,3},B={2,3,4}A={1,2,3},B={2,3,4}?
Based on these examples, you should be able to make a good guess at what the answer should be. Now, try to prove it! (As usual, this will look like "Assume x∈A−Bx∈A−B. Then [stuff]. So x∈B−Ax∈B−A. etc.")
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