Are there any numbers with 2, 3, 7 or 8 in the units place that have rational square roots?

What is "2, 3, 7, or 8 in the units"? Do you mean as a digit? 121. Do you mean as a prime factor? 4. Something else? Dunno.

I think you mean is any root of x other then actual squares (4, 9, 16 etc.) create a rational number, and the answer is no. All numbers that arent perfect squares produce irrational numbers when square rooted.

I'm pretty sure the op meant is there any perfect square with a 2, 3, 7, or 8 in the units place a.k.a. the ones place.

 

As for the question itself, imagine doing the calculation by the "long multiplication" method. Consider doing it with both numbers being equal.

 

We can agree that the units digit of the result must be equal to the units digit of the product of the units digit of the number being squared. Since there are only 10 possible digits in the units place of the number to be squared, there are a finite amount of possiblities to consider. Namely, the squares of the 10 digits, namely: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81. The units digits of these numbers are only 0, 1, 4, 5, 6, and 9. Thus showing that there are no perfect squares with a units digit of 2, 3, 7, or 8.

 

Hope this helps.

Thanks Cosine. That's what I needed.

Are there any numbers with 2, 3, 7 or 8 in the units place that have rational square roots?

 

Yes there are:

 

[math]\sqrt{2.25} = 1.5[/math]

 

[math]\sqrt{3.24} = 1.8[/math]

 

[math]\sqrt{7.29} = 2.7[/math]

 

[math]\sqrt{8.41} = 2.9[/math]

 

Though you may not have worded yuor question correctly if you are after natural numbers.

I think the implication was probably natural numbers.