Maths-
General
Easy

Question

Show that  square root of 3-square root of 2  is an irrational number.

hintHint:

Hint:
The real numbers which cannot be expressed in the form of p/q(where p and q are integers and q ≠ 0) are known as irrational numbers. In the given question we are asked to prove if square root of 3 and  square root of 2 is irrational number. To do so we use the method of contradiction.

The correct answer is: Is an irrational number


    Let's assume that square root of 3-square root of 2 is a rational number.
    Step 1 of 2:
    ¶If square root of 3 space end root-square root of 2 is rational, that means it can be written in the form of p/q, where p and q integers q ≠ 0.
    ¶So, square root of 3 minus square root of 2 equals p over q
    ¶Squaring both sides
      So, left parenthesis square root of 3 minus square root of 2 right parenthesis squared equals p squared over q squared
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell 3 plus 2 minus 2 cross times square root of 3 cross times square root of 2 equals p squared over q squared end cell row cell 5 minus 2 square root of 6 equals p squared over q squared end cell row cell 5 minus p squared over q squared equals 2 square root of 6 end cell row cell fraction numerator 5 q squared minus p squared over denominator 2 q squared end fraction equals square root of 6 end cell end table
    Step 2 of 2:
    ¶Here, we see that fraction numerator 5 q squared minus p squared over denominator 2 q squared end fraction is a rational number. But  square root of 6 is an irrational number. So our assumption is wrong.
    Final Answer:
    ¶As square root of 6 is an irrational number. So fraction numerator 5 q squared minus p squared over denominator 2 q squared end fraction  must also be an irrational number. Thus, our assumption is wrong. Hence, square root of 3 minus square root of 2 is an irrational number

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