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# is the square root of ten rational or irrational

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is the square root of ten rational or irrational

Jan 10, 2015

#1
+27377
+10

Irrational

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Jan 10, 2015

#1
+27377
+10

Irrational

.

Alan Jan 10, 2015
#2
+95361
+5

Lets try and prove this.  I am going to do it by contradiction.

$$\\Assume \;\;\sqrt{10}\; is\; rational\\\\ then \;\sqrt{10}\;can\;\; be\;\; written\;\; as \;\;\dfrac{p}{q}\\ where p and q are relatively prime integers (they have no common factors)\\\\$$

$$\\\sqrt{10}=\frac{p}{q}\\\\ 10=\frac{p^2}{q^2}\\\\ p^2=10q^2\\\\ So\;\; p^2\;\; is a multiple of 10\\\\ \mbox{so p is a multiple of 10}\\\\ Let\; p=10g\\\\ (10g)^2=10q^2\\\\ 100g^2=10q^2\\\\ 10g^2=q^2\\\\ So\;\; q^2  is a multiple of 10\\\\ so q is a multiple of 10\\\\ I have discovered that p and q are both multiples of 10$$\$

Therefore p and q are not relatively prime

therefore the original statement is contradicted

therefore  $${\sqrt{{\mathtt{10}}}}$$    is irrational.

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Note:

I have said that since pis a multiple of 10, p must also be a multiple of 10. WHY is so.

Well the prime factors of squared numbers must come in pairs.

Lets look at an example

$$\\What\;\; if\;\; p^2=900\\ p^2=30\times 30\\ p^2=3\times 5\times 2 \quad \times \quad 3\times 5\times 2\\ p^2=3\times 3\times 2 \times 2\times 5\times 5\\ see how the factors have to be in pairs?\\ p=3\times 2 \times 5\\$$

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10 is the product of 2 prime numbers

Can you see how if  p^2 is a product of 10 then p must also be the product of 10?

Jan 11, 2015