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jadahilbun01
19.08.2020 •
Mathematics
Daniel says that when the irrational number 7√3 is multiplied by any rational number, the product is always an irrational number. What value for the rational number disproves Daniel's claim?
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Ответ:
0
Explanation:
When we multiply 0 by any number, we get 0 as a result
x*0 = 0
0*x = 0
for any number x.
The number 0 is rational since we can write it as a fraction of two integers
0 = 0/1
If Daniel were to correct his statement to say "multiply by any nonzero rational number", then his statement would be correct that the result is irrational.
Extra info:
Here's a proof showing why Daniel's claim is correct if we consider nonzero rational numbers
Let p be a nonzero rational number, so p = a/b for integers a,b where neither a or b are zero
Let q be an irrational number. We cannot write q as a ratio of two integers
The claim is that p*q is irrational. For now let's assume the opposite. So assume p*q is rational. This means p*q = r/s for integers r,s
This would be the same as (a/b)*q = r/s which solves to q = (r/s)*(b/a) = (rb)/(sa) making q rational, but that contradicts the fact we made q irrational earlier.
Therefore, the assumption p*q is rational cannot be the case, and p*q must be irrational.
Ответ:
4.3722893i4u37738jdjd8ddjdijdej