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?

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.

answer:

it will cost 480 dollars to refurbish the larger wood floor

step-by-step explanation:

as it's a ratio of 4 that costs 384 dollars, divide the price by the ratio

384 / 4 = 96 dollars

96 dollars per unit of the ratio

now to get the price to refurbish the larger wood floor, times the price by the units

96 * 5 = 480

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