On Bradley's level?
I don't think Floyd has ever done it.
Canelo, no
Guererro, please
Cotto, top fighter but after the beatings he took from Pacquiao and
Margarito he was someway past his prime.
Ortiz, don't make me laugh.
Mosley, about 7 years or more past his prime.
Marquez, past prime and bloated at the weight having jump 3
weights in a year at the age of 35.
Hatton, prime but not on Bradley's level.
Oscar, part time boxer.
Baldomir, don't be so stupid.
Has Floyd ever beaten a Prime fighter
posted on 15/4/14
Lol probably since Chico never even fought above lightweight I think, it's as silly as saying ward would batter Bradley who is 2 weights naturally below him
posted on 15/4/14
Or 3 weights
posted on 15/4/14
Bradley fought at light welter and so did Corrales.
And people are mentioning Canelo vs Bradley.
Moving the goalposts just like Floyd
posted on 15/4/14
You come out with the same tired things after every pac or may fight. After every may win, you discredit his opponent. After every pqc win you big up his opponent as some great fighter. Even when their opponent is the same fighter. Give it a rest man.
posted on 15/4/14
You and everyone else knows the answer is no.
Then the thought dawns on you that Bradley is not that great, then you have to ask yourself why has Floyd been able to get away with it for so long.
posted on 18/4/14
He has built his image and reputation on the 0.
He's a great champion, but has avoided taking risks in his choice of opponent.
His next opponent is a perfect example of this, Maidana style wise is tailor made for Floyd.
posted on 19/4/14
Meh Pacs just the best of the " rest ", Mayweathers in his own league. Sadly he'll never get the respect he deserves until he's dead
posted on 19/4/14
Manny Pacquiao is the fighter of the decade, voted for by the American Boxing Writers Association.
posted on 21/4/14
No. It's why he isn't one of the goats, you can win all the fights you want but if none of them are a high level of fighter it's not worth much.
posted on 21/4/14
Wlad is another example of this.