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Is it possible to multiply by 'undefined'? 
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#1
Dec304, 03:46 PM

P: 21

My trig teacher says that my solution would not work because you can't multiply by undefined. Why can't you? Is there a rule, postulate, etc... that says you can't? If so, what is it? I think that if you multiply by undefined you would get undefined. Is this wrong.
Thanks for reading. 


#2
Dec304, 05:06 PM

P: 48

Try to think of exact examples. 1/0 would be undefined. Lets assume that algebraic rules do apply to such numbers, and you can multiply by them. What happens if you multiply 1/0 by 0? The 0's would cancel out and you would get 1 as an answer. First of all this is not undefined, as you speculated, but there is also another problem: What happens if you multiply this by 2? 1*2=2, obviously. One property of multiplication is that it has to be commutative, in other words the order in which you perform multiplications does not effect the answer. a*b=b*a and similarly (a*b)*c=a*(b*c).
Apply this to the above example. As we showed (1/0*0)*2=(1)*2=2 but what happens if we perform the multiplications in a different order? 1/0*(0*2)=1/0*0=1 These answers are not the same, so normal multiplication does not apply to undefined numbers. There are many other such contradictions that can be found. It's similar to the problem of performing algebraic operations on infinity as if it were a number. I'm sure this could have been said much more elegantly, but I'm tired right now. Did this answer your question? 


#3
Dec304, 07:35 PM

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#4
Dec404, 08:43 PM

P: 21

Is it possible to multiply by 'undefined'?
Eddo, you said, "What happens if you multiply 1/0 by 0? The 0's would cancel out and you would get 1 as an answer." That is not true. 1/0*0 is not 1. When you multiply a fraction by a whole number, you change the second into a fraction:
1/0*0/1=0/0 0/0 is also undefined; therefore, even 1/0*0 is undefined. So you are wrong. Tide, you asked me what the point would be? The point is that the teacher said that the only completely correct answer was wrong. This miss leads the class. The question was: Tan(x)Sec(x)=Tan(x) What is the general solution for x? My solution was: Sec(x)=Tan(x)/Tan(x) (When you divide anything by itself, you get one of two solutions: +1 and "undefined", s= Sec(x)=+1 or Sec(x)=undef. x=360n x=90+180n But, in order for Tan(x)/Tan(x) to equal undef.: Tan(x)=undef. or Tan(x)=0 x=90+180n x=180n So, this yeilds three general solutions: {x=360n,x=90+180n,x=180n} This can all be condensed into: {x=90n} 


#5
Dec404, 11:33 PM

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Moore, "undefined" is not a quantity. You don't treat is as a quantity on one side or other of an equation.
It is a quality. Something is undefined if it is not defined within some ring. 


#6
Dec504, 09:35 AM

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You need to be careful telling others they are wrong. Eddo carefully qaulified his statement by saying in effect what if we treat 1/0 as any other real number, and pretend that multiplication worked. Of course 1/0 isn't a real number, so the premise is false, but that isn't the important thing there. In the real numbers (a/b)*b=a for all a when b is not zero. That is what he was getting at in attempting to explain what "undefined" might mean. 


#7
Dec504, 02:59 PM

P: 48

Thank you Matt, saved me having to defend myself.
I don't know if you know this already or not Moore, but if you wanted to solve the problem in order to avoid any "undefined"s simply rearrange it so that one side is equal to zero: tanxsecx=tanx tanxsecx  tanx=0 tanx(secx  1)=0 so tanx=0 or secx=1 


#8
Jul508, 08:13 PM

P: 143

It uses the special value NULL for all types, which would be equal to 'Undefined'. Forexample: NULL or TRUE = TRUE NULL or FALSE = NULL NULL and FALSE = FALSE 1 + NULL = NULL 'A' + NULL = 'A' (string concatenate, NULL is them emtpty string) 


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