Home > categories > Chemical > Additives > In Linear Algebra, how can you show that there is no additive identity in (x1,y1) + (x2,y2) (0,0)?
Question:

In Linear Algebra, how can you show that there is no additive identity in (x1,y1) + (x2,y2) (0,0)?

Let V be the set of ordered pairs (x,y) of real numbers with addition and scalar multiplication defined by:(a). (x1, y1) + (x2,y2) (0,0)(b). K(x,y) (Kx, Ky).i). Show that there is no additive identity.ii). Show that every element of V has more than one inverse.iii). Does V qualify to be a vector space?

Answer:

The additive identity (e) is defined as: x + e x x - e
Yes it is unsafe. If only using the front breaks on a bike you can more likely flip the bike forward if the stop is to fast. It is best to either wear pants or hike your dress up. You should always use the intended safety methods for any bike or car to ensure yours and others safety.
Lets see: 1. Off balance: CHECK. 2. Loose clothing: CHECK 3. Exposed skin: CHECK 4. No rear brake: CHECK 5. No option to stop weighted on right side: CHECK 6. No use of legs to hold/control bike balance: CHECK I'm all gear, all the time, so your idea sounds terrible to me. But it is your body/bike and it is your decision.
it may take a little bit of skill, which I believe you can master. Besides, I?ve been riding without rear brakes for about a year., and if you?re not that fast while riding, you should be ok.
An additive identity would have to satisfy (x,y)+e(x, y) which includes as a special case (1,0)+e(1, 0). But (1, 0) + Anything (0, 0). So there is no additive identity. As to part ii: I do not see how an inverse can be defined if there is no additive identity in the first place. Usual definition of inverse of a vector v is another vector u such that v+u additive identity. Ask your teacher about this!

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