Question

prove 11root7 is irrational​

Answer 1

Answer:

√7=a/b ( on squaring both sides). Here 7 divides a² so for it will also divide a . Here 7 divides b² so for 7 divides b. Here we find 7 is common which divide both a and b but this is contradiction because a and b are co prime they don't have common factor other than 1.

Answer 2

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√11+√7

√11+√7spoz a= √11+√7

√11+√7spoz a= √11+√7a= √11+√7

√11+√7spoz a= √11+√7a= √11+√7a-√11=√7

√11+√7spoz a= √11+√7a= √11+√7a-√11=√7(a-√11)^2=(√7)^2

√11+√7spoz a= √11+√7a= √11+√7a-√11=√7(a-√11)^2=(√7)^2a^2-2√11a+11^2=49

√11+√7spoz a= √11+√7a= √11+√7a-√11=√7(a-√11)^2=(√7)^2a^2-2√11a+11^2=49a^2 -2√11a+121 = 49

√11+√7spoz a= √11+√7a= √11+√7a-√11=√7(a-√11)^2=(√7)^2a^2-2√11a+11^2=49a^2 -2√11a+121 = 49a^2-2√11a+121-49=0

√11+√7spoz a= √11+√7a= √11+√7a-√11=√7(a-√11)^2=(√7)^2a^2-2√11a+11^2=49a^2 -2√11a+121 = 49a^2-2√11a+121-49=0a^2-2√11a+72=0

√11+√7spoz a= √11+√7a= √11+√7a-√11=√7(a-√11)^2=(√7)^2a^2-2√11a+11^2=49a^2 -2√11a+121 = 49a^2-2√11a+121-49=0a^2-2√11a+72=0but √11 is irrational number

√11+√7spoz a= √11+√7a= √11+√7a-√11=√7(a-√11)^2=(√7)^2a^2-2√11a+11^2=49a^2 -2√11a+121 = 49a^2-2√11a+121-49=0a^2-2√11a+72=0but √11 is irrational numberso √11+√7 is an irrational number..

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