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Question Number 12822 by fawadalamawan@gmail.com last updated on 02/May/17

$$provebycontradiction9+13\sqrt{3}$$$$isirrational$$

Answered by mrW1 last updated on 03/May/17

$$letusassume9+13\sqrt{3}isrational,.i.e.$$$$thereexistintegernumbersaandb,b\ne 0,$$$$9+13\sqrt{3}=\frac{a}{b}$$$$\Rightarrow \sqrt{3}=\frac{a-9b}{13}=\frac{integer}{integer}=rational$$$$thatistosay:if9+13\sqrt{3}isassumed$$$$toberational,then\sqrt{3}isalsorational,$$$$butthisisnottrue,therefore$$$$9+13\sqrt{3}isirrational.$$

Commented by Joel577 last updated on 04/May/17

$$howtoprovethat\sqrt{3}isirrational?$$

Commented by mrW1 last updated on 04/May/17

$$assume\sqrt{3}wererational,i.e.$$$$\sqrt{3}=\frac{a}{b},and\frac{a}{b}cannotbefurther$$$$simplified,i.e.aandbehaveno$$$$commonfactor.$$$$$$$$3=\frac{a^2}{b^2}$$$$a^2=3b^2$$$$$$$$ifbiseven,i.e.b=2n$$$$a^2=3\times 4n^2=even$$$$\Rightarrow a=even=2m$$$$\Rightarrow aandbhavecommonfactor2,$$$$\frac{a}{b}canbefurthersimplifiedto\frac{m}{n},$$$$\Rightarrow bisnoteven!$$$$$$$$ifbisodd,i.e.b=2n+1$$$$b^{2}isalsoodd,3b^{2}isalsoodd,i.e.$$$$a^{2}isodd,amustbealsoadd,i.e.a=2m+1$$$${(2m+1)}^2=3{(2n+1)}^2$$$$4m^2+4m+1=12n^2+12n+3$$$$2(m^2+m)=6(n^2+n)+1$$$$even=odd$$$$\Rightarrow bisnotodd.$$$$$$$$\Rightarrow thereexistnointegeraandbtofulfill\sqrt{3}=\frac{a}{b},$$$$\Rightarrow \sqrt{3}isirrational$$

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