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So we all know that if you take a fraction, say 2/3, and multiply the top and bottom both by the same term, the fraction is still going to equal 2/3, right? So say we multiply both the numerator and denominator by 0, wouldn’t we get the undefined 0/0? Or would we solve this exactly as we would if we subbed 0 into 2x/3x? If it was solved that way, then it would make sense for it to still equal 2/3 as it should. would 0 be treated almost as a removable discontinuity in this case. or would we treat it as 2(0)/3(0)=0/0=undef?

- kristinabui8
**Posts:**2**Joined:**Thu Jul 30, 2020 4:54 am**Reputation:****1**

I think you are not quite understanding what you are saying!

Almost! If you multiply both top and bottom by any number other than 0 the fraction will still equal 2/3.

Ah, well that's the situation I just disallowed isn't it! Yes, instead of 2/3 it would now be "undefined". And that is because "multiplying the bottom of a fraction by 0" is the same as "dividing by 0" which is disallowed operation.

This is where you make your main mistake. If we subbed 0 into 2x/3x it would NOT "still equal 2/3" nor is there any reason to say "as it should". If we sub x= 0 into 2x/3x we get "undefined".

No, 0 would not be treated as a "removable discontinuity" but you might just be saying it incorrectly. It isn't the number 0 that is a "removable discontinuity". It is rather a point on (or rather not on) the graph of y= 2x/3x. The graph of y= 2/3 is a horizontal straight line. Since 2x/3x= 2/3 for all x except x= 0, the graph of y= 2x/3x is that same straight line except that there is a "hole" in it- the point (0, 2/3) is missing.

kristinabui8 wrote:So we all know that if you take a fraction, say 2/3, and multiply the top and bottom both by the same term, the fraction is still going to equal 2/3, right?

Almost! If you multiply both top and bottom by any number other than 0 the fraction will still equal 2/3.

So say we multiply both the numerator and denominator by 0, wouldn’t we get the undefined 0/0?

Ah, well that's the situation I just disallowed isn't it! Yes, instead of 2/3 it would now be "undefined". And that is because "multiplying the bottom of a fraction by 0" is the same as "dividing by 0" which is disallowed operation.

Or would we solve this exactly as we would if we subbed 0 into 2x/3x? If it was solved that way, then it would make sense for it to still equal 2/3 as it should. would 0 be treated almost as a removable discontinuity in this case. or would we treat it as 2(0)/3(0)=0/0=undef?

This is where you make your main mistake. If we subbed 0 into 2x/3x it would NOT "still equal 2/3" nor is there any reason to say "as it should". If we sub x= 0 into 2x/3x we get "undefined".

No, 0 would not be treated as a "removable discontinuity" but you might just be saying it incorrectly. It isn't the number 0 that is a "removable discontinuity". It is rather a point on (or rather not on) the graph of y= 2x/3x. The graph of y= 2/3 is a horizontal straight line. Since 2x/3x= 2/3 for all x except x= 0, the graph of y= 2x/3x is that same straight line except that there is a "hole" in it- the point (0, 2/3) is missing.

- HallsofIvy
**Posts:**263**Joined:**Sat Mar 02, 2019 9:45 am**Reputation:****98**

kristinabui8 wrote:So we all know that if you take a fraction, say 2/3, and multiply the top and bottom both by the same term, the fraction is still going to equal 2/3, right? So say we multiply both the numerator and denominator by 0, wouldn’t we get the undefined 0/0? Or would we solve this exactly as VidMate Mobdro we would if we subbed 0 into 2x/3x? If it was solved that way, then it would make sense for it to still equal 2/3 as it should. would 0 be treated almost as a removable discontinuity in this case. or would we treat it as 2(0)/3(0)=0/0=undef?

You’re not “multiplying the top and bottom by x”, you’re multiplying the whole fraction by 1=x/x, which is only defined for x≠0, as others have noted.

- kristinabui8
**Posts:**2**Joined:**Thu Jul 30, 2020 4:54 am**Reputation:****1**

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