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When a nonzero rational and an irrational number are multiplied, is the product rational or irrational? Explain.

Guest Jun 18, 2015

#3
+27219
+18

Suppose the rational number is m/n, where m and n are integers.

Let the irrational number be t

Assume the product t*m/n is rational; then it could be represented by a/b, where a and b are integers.

So we would have a/b = t*m/n

Multiply both sides by n/m to get:  t = a*n/(b*m)

a and n are integers so a*n = p, another integer.

Similarly, b*m = q, an integer.

But this means that t = p/q, the ratio of two integers.  In other words, t, our irrational number turns out to be rational!

This is a contradiction of course, which means that our initial assumption, that an irrational number multiplied by a rational number is rational, is false.

In other words, the product of an irrational number and a rational number is irrational.

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Alan  Jun 18, 2015
#1
+92367
+5

Irrational........

Think about this   .......if we took  √17  (an irrational) and added it four times, we would just have 4√17.....but the "√17"  part would remain, and thus, the "irrational" part would remain......the only way to make this "rational"  would be to multiply it by an irrational!!!

CPhill  Jun 18, 2015
#2
+223
+10

but...

a rational number is a integer number or a fraction of two integers.

an irrational number is a number that can not be written as a fraction. (pi, sqrt(2), e for example)

therefore, when you multiply an irrational number with a rational number it has to be irrational (if the rational number isn´t zero)

for example multiplie pi with 1/4 the result is pi/4 but since you can´t write pi as a fraction you can´t write pi/4 as a fraction.

#3
+27219
+18

Suppose the rational number is m/n, where m and n are integers.

Let the irrational number be t

Assume the product t*m/n is rational; then it could be represented by a/b, where a and b are integers.

So we would have a/b = t*m/n

Multiply both sides by n/m to get:  t = a*n/(b*m)

a and n are integers so a*n = p, another integer.

Similarly, b*m = q, an integer.

But this means that t = p/q, the ratio of two integers.  In other words, t, our irrational number turns out to be rational!

This is a contradiction of course, which means that our initial assumption, that an irrational number multiplied by a rational number is rational, is false.

In other words, the product of an irrational number and a rational number is irrational.

.

Alan  Jun 18, 2015