+0

0
397
3

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

Guest Jun 18, 2015

#3
+26399
+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
Sort:

#1
+80935
+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
+26399
+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

19 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details